
Class _lU_il5_ 

Book._ .K 44 

(kpightF 



CDPffilGHT DEPOSnv 



I 



DESIGN 



OF A 



High-Speed Steam Engine 

Notes, Diagrams, Formulas and Tables. 



SECOND EDITION, 

REVISED AND ENLARGED. 



BY 

J. F. KLEIN, 

Professor of Mechanical Engineering in Lehigh University^ 



,'\ ^' \.' 



BETHLEHEM, PA.: 

Times Publishing Company, 

1903. 






THE LIBRARY OF 

CONGRESS, 

One Copy Received 

FEB. M 1903 

COPVWQHT ENTWV 

CLASS ^^-XXol No 
COPY 8. 



Copyright 1903, 
BY J. F. Klein. 



t-0 



^^k' 



PREFACE. 

When the first edition of this book was pubHshed the intention 
was to make it Part I of a treatise on the Dynamics and Design 
of a High-Speed Steam Engine. Part II was to treat of the 
Dynamics and Design of Shaft Governors. This second part 
is now nearly finished, but proves to be so difficult and volumi- 
nous as to make it unsuitable for most undergraduate work in 
technical schools ; it will, therefore, be issued as a separate 
and independent work. 

The present volume drops its old designation as Part I, corrects 
the typographical errors of the first edition, and adds much new 
material in the form of appendices. 

J. F. Klein. 

Bethlehem, Pa., February, IQOJ. 



NOTES, DIAGRAMS, FORMULAS AND TABLES 



FOR THE DESIGN OF A 



HIGH-SPEED STEAM ENGINE 



Data. 

Determine the principal dimensions of an engine fulfilling the 
following requirements : 

The work ordinarily required, = lOO H, P. net = i lO /. H. P., 
but the load may vary from 50 /. H. P. to 170 /. H. P. 

The engine is to run steadily under all these loads, the total 
variation (N^ — N^ of the number of revolutions of engine shaft 
being less than Jtr oi the normal number of revolutions {N) 

50 2 

The engine is to be of the non-condensing, high (rotative) 
speed type, durable, of compact and rigid form, and the steam 
consumption of the 100 H.P. per hour is to be a minimum. 
Moreover, this high-speed engine must be provided with gen- 
erous wearing surfaces which are continuously lubricated, the 
reciprocating parts must be balanced and their weights so chosen 
that the tangential pressures on the crank pin will be as nearly 
constant as possible. 



ORDER OF CALCULATION. 



It is assumed that the workmanship will be excellent, the par- 
allelism of crank pin and shaft, as well as the alignment of crank- 
shaft bearings, perfect, and that they will be maintained so by- 
good foundations. 



Order of Calculation. 

I. Determination of diagram of effective steam pressures 
on piston of required engine. 

a. Determination of boiler pressure. 

b. Determination of clearance, real cut-off and apparent 
cut-off. 

c. Determination of back pressure and amount of com- 
pression. 

d. Method of constructing equilateral hyperbola and of 
determining points on the curves pv''^ = constant by 
tabulated coordinates. 

e. Methods of determinining the mean effective pressures 
of indicator diagrams. 

f. Determination of the cut-off corresponding to a given 
horse-power N^ when the mean effective pressure /^^^ and 
the total pressure /''^ is known for another horse-power 
H of the same engine running at the same speed, with 
same initial pressure /i and same counterpressures. 

£•. Diagrams of effective-steam-pressure-on-piston for for- 
ward and return strokes. 
II. Determination of ratio of length of connecting rod to 
that of crank. 
III. Determination of that mean accelerating force (per □'' of 
piston) necessary to start reciprocrating parts and cor- 
responding to the most uniform tangential pressure on 
crank pin. 
a. Determination of acceleration and construction of dia- 
grams of accelerating forces. 



ORDER OF CALCULATION. 3 

b. Construction of horizontal or resultant pressures on 
cross-head pin (or on crank pin when friction is neg- 
lected) by combining diagram of effective steam pres- 
sures on piston with diagram of acceleration. 

c. Discussion of effects on this diagram of valve setting, 
high speed and long and short cut-offs. 

d. Conversion of horizontal pressures on crank pin into 
tangential pressures. 

e. Construction of diagrams of tangential pressures on 
crank pin. 

IV. Determination of diameter of cylinder, length of stroke, 
revolutions per minute and weight of reciprocating parts. 
V. Preliminary estimate of dimensions of reciprocating parts, 
to see whether minimum weight of latter does not exceed 
that necessary to produce steadiness of running under 
the conditions assumed. 
VI. Determination of weight of fly-wheel rim from tangen- 
tial pressure diagram. 
VII. Determination of dimensions of crank disk and distri- 
bution of its material so that both crank arm and recip- 
rocating parts will be balanced. 
VIII. Method of determining the influence, on the foregoing 
results, of the frictional resistances, of the weight of the 
rod and of the exact values of the forces of inertia. 
IX. Calculation of width and diameter of belt pulley. 
X. Graphical determination of diameter of crank shaft. 
XI. Calculation of the length of crank shaft journal and 
graphical determination of the plane of division of its 
brasses. 
XII. Calculation of dimensions of steam ports and pipes, also 
thickness of cylinder walls. 

XIII. Valve diagrams and dimensions of valve and gear. 

XIV. Drawing of details of engine. 

XV. Drawing of plan and elevation of complete engine. 



4 determination of boiler pressure. 

Plates to be Drawn. 
I. Six Indicator Diagrams. 
II. Six Diagrams of effective steam pressures on piston. 

III. Diagrams of accelerating and retarding pressures by- 
approximate and exact methods. 

IV. Six Diagrams of horizontal pressures on cross-head pin. 
V. Diagram for converting horizontal pressures on cross- 
head pin into tangential pressures on crank pin. 

VI. Six Diagrams of tangential pressures on crank pin. 

VII. Four Diagrams of the forces shaking engine bed, for as 
many different degrees of counter-weighting. 
VIII. Distribution of material on crank disk, and determina- 
tion of its center of gravity. 

IX. Determination of diameter of engine shaft by the methods 

of Graphical Statics. 
X. Graphical determination of the direction of the resultant 
pressure on crank-shaft bearing. 

XL Valve Diagrams. 

Plates containing details, plan and elevations of com- 
plete engine. 

I A. 

Determination of Boiler Pressure. 

If we consider the cylinders as perfect non-conductors of heat, 
and consider economy of steam only, we must, in accordance 
with the deductions of thermodynamics, make our initial steam 
pressures as high as possible, for by so doing we increase the 
range of temperature of the steam in the engine, and thereby 
increase the efficiency of the engine. This is clearly shown by 
the expression for the thermal efficiency rj of the engine 

T T 

7*1 being the absolute temperature of the steam at the initial 
absolute pressure p^, and T^ the lowest absolute temperature 



DETERMINATION OF BOILER PRESSURE. 5 

practicably possible for the engine, being in non-condensing 
engines 7^2= 212° + 459.°4 ^ 6ji°4 Fahr. T^i increases very 
slowly with the higher pressures of steam, see Roentgen's Ther- 
modynamics, pp. 672 and 674. 

Even for ordinary (cast iron) conducting cylinders it is true 
(up to not very well defined limits) that economy in the use of 
steam increases slowly with the initial pressure. Thus according 
to Prof. Thurston the weight of steam per hour per H. P. should 
in good engines be not more than that given by the empirical 
formula 

200 

^ pounds (3) 

and for the best practice with large engines, dry steam, high 
piston speed and good design, construction and management the 
consumption of steam should be not more than 

150 

In (3) and (4) /i is the absolute pressure in pounds per sq. inch. 

Sometimes the highest pressure which can be employed in an 
engine is prescribed by the strength of the boilers which are to 
furnish it with steam ; at other times (with engines running under 
conditions in which the power occasionally sinks considerably 
below the average) a lower pressure (accompanied by a greater 
cut-off) than economy of fuel and steam would prescribe, must 
be employed in order that the engine may run at high speed with 
great steadiness and without reversing the direction of the hori- 
zontal pressures on crank pin. See limits to high speed. III, c. 

The employment of very high pressures is also limited by the 
greater first cost of the boilers owing to the greater strength 
required, and the greater care which must be exercised in their 
construction. The engine also becomes somewhat more costly 
for the same reason. A practical limit is thus set to the general 
use of extremely high pressures. 



6 CLEARANCE. 

In the present case we will start the calculation by assuming 
the initial boiler pressure employed by many makers of high- 
speed engines when designing and rating their engines, namely, 
85 pounds gauge pressure. 

I B. 

Determination of Clearance, Real Cut-off and Apparent 

Cut-off. 

Clearance is the volume or space included between the piston 
(when the latter is at the end of its stroke) and the cylinder head, 
and also includes the steam passages. It is usually expressed as 
a fraction of the volume swept through by the piston, i.e., if 2 repre- 
sents the clearance, A the area of piston and 5 the stroke we have 

clearance volume /^x 

' = AS ^5) 

The clearance should be as small as possible, for when it is at 
all large it involves considerable loss by presenting a large space 
to be filled by fresh steam, which steam is not engaged in push- 
ing back the piston when entering the cylinder, but in giving 
whirling motions to its own particles, though during expansion 
a portion of the energy of the steam which has filled the clear- 
ance spaces is utilized. On account of the loss occasioned by it, 
the clearance is sometimes called the hurtful space. Its hurtful 
influence can, however, be entirely removed by allowing the ex- 
haust steam to be compressed till it reaches a pressure equal to 
the initial pressure. When this is done no fresh steam is needed 
to fill the clearance spaces at the beginning of the stroke ; but as 
it is not always practicable (as will appear later) to compress the 
exhaust steam to so great an extent, it behooves the designer to 
diminish the clearance as much as possible. 

The total amount of engine clearance employed in practice 
varies greatly with the type and size of engine employed. For 
example, in Corliss Engines the four separate valves employed 



REAL CUT-OFF. / 

are placed close to the cylinder and the distance between end 
of piston and cylinder head left for variation of length of con- 
necting rod (when the wear is taken up) is frequently not more 
than }('\ so that in these engines the clearance often does not 
exceed ^ of one per cent. In high-speed engines, however, the 
steam passages are usually very large, as it is very important to 
admit the steam promptly, and at full pressure. In these engines 
the clearance sometimes amounts to 14 per cent., but this is very 
large. We will for the present assume that the clearance in our 
high-speed engine is as great as 7 per cent. 

Cut-off. 
See Fig. 3, p. 16. 
The volume of steam in the cylinder (on the live steam side of 
the piston) when the admission valve closes, divided by the vol- 
ume of this steam (after it has expanded) when the exhaust valve 

opens, is called the real or virtual cut-off =- == e. It is always 

a proper fraction. The reciprocal of this fraction is the expan- 
sion, and is of course a whole or mixed number. 

The distance of the piston from the end of its stroke when the 
admission valve closes, divided by the whole stroke, is the appar- 

e7it cut-off =r — = £\ The relation between the real and apparent 

cut-off is represented by the following equation: 

^ 4 iSA i -f i 

real cut-off = - := o-^r~; — ~c^i = = ^- (6) 

r SA ^ iSA I -f 2 ^ ^ 

The following tables, pp. 8 and 9, give this relation, and were 
taken from Du Bois' translation of Weisbach, Vol. II. If our cyl- 
inders were non-conductors, and economy of steam were the only 
element of economy to be considered, we know from thermo- 
dynamics that it would be wise to employ both high pressures 
and great expansions. Unfortunately our cylinders are con- 
ductors of heat, and as a consequence there is a very marked 



REAL CUT-OFF. 



TABLE I. 

Real Cut-off Corresponding to Apparent Cut-off for 
Different Fractions of Clearance. 



§? 










Clearance. 










S5C 
u 


•05 




















"3 A 
.05 


.01 


.02 


•03 


.04 


•05 
.095 


.06 
104 


.07 
.112 


.08 
.120 


09 
.128 


.10 
.136 


.059 


.069 


.078 


.C87 


.06 


.069 


.078 


.087 


.096 


.105 


113 


.122 


.130 


.138 


.145 


.06 


.07 


.079 


.088 


.097 


.106 


.114 


.123 


•131 


•139 


.147 


•155 


.07 


.08 


.089 


.098 


.107 


.115 


.124 


.132 


.140 


.148 


.156 


.164 


.08 


.09 


.099 


.108 


.117 


.125 


•133 


.142 


.150 


•157 


.165 


• 173 


.09 


.10 


.109 


.118 


.126 


•135 


• T43 


.151 


•159 


.167 


.174 


.182 


.10 


.11 


.119 


.127 


•136 


.144 


.152 


.160 


.168 


.176 


.183 


.191 


.11 


.12 


.129 


.137 


.146 


•154 


.162 


.170 


.178 


.185 


•193 


.200 


.12 


.13 


.139 


.147 


•155 


.163 


.171 


.179 


.187 


.194 


.202 


.209 


•13 


.14 


.149 


•157 


.165 


• 173 


.181 


.189 


.196 


.204 


.211 


.218 


.14 


.15 


.158 


.167 


.175 


.183 


.190 


.198 


.206 


.214 


.220 


.227 


•15 


.16 


.168 


.176 


.184 


.192 


.200 


.208 


.215 


.222 


.229 


•236 


.16 


.17 


.178 


.186 


.194 


.202 


.210 


.217 


.224 


.231 


.239 


•245 


•17 


.18 


.188 


.196 


.204 


.212 


.219 


.226 


.234 


.241 


.248 


•255 


.18 


•19 


.198 


.206 


.214 


.221 


.229 


.236 


•243 


.250 


.257 


.264 


.19 


.20 


.208 


.216 


.223 


.231 


•238 


•245 


.252 


•259 


.266 


.273 


.20 


.21 


.218 


.225 


.233 


.240 


.248 


•255 


.262 


.269 


•275 


.282 


.21 


.22 


.228 


•235 


.243 


.250 


.257 


.264 


.271 


.278 


.284 


.291 


.22 


.23 


.238 


•245 


.252 


.260 


.267 


.274 


.280 


.287 


•294 


.300 


•23 


.24 


.248 


•255 


.262 


.269 


.276 


.283 


.290 


.296 


•303 


•309 


.24 


.25 


.257 


.265 


.272 


.279 


.286 


.292 


.299 


.306 


.312 


•318 


•25 


.26 


.267 


.275 


.282 


.288 


•295 


.302 


.308 


•315 


.321 


.327 


.26 


.27 


.277 


.284 


.291 


.298 


•305 


•311 


•318 


.324 


•330 


•336 


.27 


.28 


.287 


.294 


.301 


.308 


•314 


.321 


•327 


•333 


•339 


•345 


.28 


.29 


.297 


•304 


•311 


•317 


•324 


•330 


•336 


•343 


.349 


•355 


.29 


.30 


.307 


•314 


•320 


• 327 


•333 


•340 


•346 


•352 


•358 


.364 


•30 


.31 


.317 


•324 


•330 


■337 


•343 


•349 


•355 


.361 


•367 


•373 


•31 


.32 


•327 


•334 


■340 


.346 


.352 


•358 


.364 


•370 


•376 


.382 


•32 


•33 


•337 


•343 


•350 


.356 


.362 


.368 


•374 


.380 


•385 


•391 


•33 


.34 


•347 


•353 


•359 


•365 


.371 


.377 


•383 


•389 


•395 


.400 


.34 


.35 


•356 


•363 


•369 


•375 


•381 


.387 


•393 


.398 


.404 


.409 


•35 


.36 


•366 


•373 


•379 


•385 


•390 


•396 


.402 


.407 


•413 


.418 


•36 


.37 


.376 


•382 


•388 


•394 


.400 


.406 


.411 


.417 


.422 


.427 


•37 


.38 


.386 


•392 


•398 


•404 


.410 


.415 


.421 


.426 


•431 


•436 


•38 


•39 


.396 


.402 


.408 


•413 


.419 


.425 


.430 


•435 


.440 


.445 


•39 


.40 


.406 


.412 


.417 


•423 


.429 


.434 


•439 


.444 


.450 


•455 


' .40 



REAL CUT-OFF. 



TABLE I (Continued). 

Real Cut-off Corresponding to Apparent Cut-off for 

Different Fractions of Clearance. 



£d 


1 




Clearance. 










gd 


u. O 
.41 
















.41 


.01 


.02 


•03 


.04 


.05 


.06 


.07 


.08 


.09 


.10 

463 


.416 


.422 


.427 


•433 


•438 


.443 


.449 


•454 


459 


.42 


.426 


•431 


437 


.442 


•448 


453 


458 


•463 


.468 


•473 


.42 


43 


•436 


.441 


.447 


.452 


457 


.462 


.467 


•472 


.477 


.482 


•43 


.44 


.446 


.451 


.456 


.462 


.467 


•472 


•477 


.481 


.486 


.491 


.44 


•45 


.455 


.461 


.466 


.471 


.476 


.481 


.486 


.491 


•495 


.500 


•45 


.46 


.465 


■^v 


•H^ 


.481 


.486 


•491 


•495 


.500 


•505 


.509 


.46 


47 


.475 


.480 


485 


.490 


•495 


.500 


•505 


.509 


•514 


.518 


•47 


48 


.485 


.490 


•495 


.500 


.505 


.509 


.514 


•519 


•523 


.527 


.48 


49 


•495 


.500 


.505 


.510 


.514 


.519 


•523 


.528 


•532 


.536 


.49 


.50 


•505 


.510 


•515 


.519 


.524 


.528 


•533 


•537 


.541 


•545 


•50 


•51 


•515 


.520 


.524 


.529 


•533 


•538 


•542 


.546 


•550 


•554 


•51 


.52 


•525 


.529 


•534 


•538 


.543 


.544 


•551 


•556 


.560 


.564 


•52 


•53 


•535 


•539 


•544 


■548 


•552 


.557 


.561 


•565 


.569 


.573 


•53 


•54 


•545 


•549 


•553 


•558 


.562 


.566 


•570 


•574 


.578 


.582 


•54 


.55 


•554 


•559 


.563 


.567 


•571 


.575 


•579 


•583 


.587 


•591 


.55 


.56 


.564 


.569 


•573 


•577 


.581 


.585 


•589 


•593 


•596 


,600 


•56 


•57 


.574 


.578 


•583 


.587 


.590 


.594 


•598 


.602 


.606 


.609 


•57 


.58 


.584 


.588 


.592 


.596 


.600 


.604 


.607 


.611 


.615 


.618 


•58 


•59 


•594 


• 598 


.602 


.606 


.610 


.613 


.617 


.620 


.624 


.627 


•59 


.60 


.604 


.608 


.612 


.615 


.619 


.623 


.626 


.630 


.633 


.636 


.60 


.61 


.614 


.618 


.621 


.625 


.629 


.632 


.636 


•639 


.642 


.645 


.61 


.62 


.624 


.627 


.631 


•635 


.638 


.642 


.645 


.648 


.651 


•655 


.62 


•63 


.634 


.637 


.641 


.644 


.648 


.651 


.654 


.657 


.661 


.664 


.63 


.64 


.644 


.647 


.650 


.654 


.657 


.660 


.664 


.667 


.670 


.673 


.64 


.65 


•653 


.657 


.660 


.663 


.667 


.669 


.673 


.676 


.679 


.682 


.65 


.66 


.663 


.667 


.670 


.673 


.676 


.679 


.682 


.685 


.688 


.691 


.66 


.67 


.673 


.676 


.680 


.683 


.686 


.689 


.692 


.694 


.697 


.700 


.67 


.68 


.683 


.685 


X89 


.692 


.695 


.698 


.701 


.704 


.706 


.709 


.68 


.69 


•693 


.695 


,e99 


.702 


.705 


.708 


.710 


.713 


.716 


.718 


.69 


.70 


•703 


.706 


.709 


.712 


.714 


.717 


.720 


.722 


.725 


.727 


.70 


.71 


.713 


.716 


.718 


.721 


.724 


.726 


.728 


.731 


•734 


.736 


.71 


.72 


.723 


.725 


.728 


■731 


.733 


.736 


.738 


.741 


.743 


.745 


.72 


.73 


.733 


•735 


•738 


•740 


.743 


•745 


.748 


.750 


•752 


•755 


•73 


.74 


.743 


•745 


.748 


.750 


.752 


•755 


•756 


•759 


.761 


.764 


.74 


•75 


.752 


.755 


•757 


.760 


.762 


.764 


.766 


•769 


.771 


•773 


•75 



lO REAL CUT-OFF. 

loss arising from the condensation of the Hve steam as it enters 
the cylinder and comes in contact with the surfaces of the latter 
which have just been cooled by contact with, and giving up their 
heat to, the exhaust steam. The complexity of this interchange 
of heat is such that thus far it has not been satisfactorily followed 
by calculation, and we must depend upon experiment for our 
knowledge of what constitutes the most economical point of 
cut-off with respect to the consumption of steam or fuel. This 
cut-off may be determined by means of an empirical formula 
representing the result of extensive and carefully conducted 
experiments by Mr. Charles Emery, namely, 



I 22 



I 

r~ J _^ A ~" 22 + /i 

22 



(7) 



e = real cut-off. r = ratio of expansion. /^ = absolute initial 
pressure. 

Mr. Emery considers that the cut-offs given by this formula 
are " nearly correct for single engines of ordinary construction, 
and too large for the better class of compound engine." The 
first two columns of Table IV on page i8 contain the initial 
absolute pressure /i for various values of e = real cut-off when 
pvji = constant represents the curve of expansion. Thus far the 
only element of economy considered in choosing the point of cut- 
off was economy of steam or of the fuel which produced it. But 
this is only one of the elements which influences the cost per 
hour of the H. P. employed. Other elements such as wages of 
attendants, interest on cost of engine and boilers, repairs on 
engine and boilers, insurance, depreciation, etc., influence the cur- 
rent cost of the //. P. employed. That cut-off which makes this 
cost a minimum is evidently the true economical point of cut- 
off, the one in which the engine owner is financially interested- 
But as this cost, per horse-power per day, does not vary much for 
a considerable range of cut-off, we will here consider only the 
principal factor in that cost, i.e., steam. 



BACK PRESSURE AND COMPRESSION. II 

We may make a sufficient approximation to the correct cut-off 
by taking a value of the latter greater than that given by Mr. 
Emery's empirical formula for greatest economy of steam or fuel, 
for it is evident from general principles that the influence of the 
other expenses will be to make the proper cut-off greater than 
that corresponding to the greatest economy of steam or fuel. 
As our initial pressure was above assumed to be 85 pounds (by 

guage) we get from Emery's formula - < — . If we now assume 

— = — we shall have, both as regards the pressure and the cut- 
off, conditions such as are assumed by many high-speed engine 
builders. 



Ic. 

Determination of Back Pressure and Amount of 
Compression. 

For well-designed non-condensing engines the back pressure 
during the exhaust is usually taken at 16 pounds (above a vacuum) 
per □'', and in condensing engines at 2 pounds per □''. As re- 
gards the amount of compression or cushioning suitable for high- 
speed engines it may be said that while on the one hand compres- 
sing up to boiler pressure has the advantage of obviating the 
necessity of partly filling the clearance spaces with fresh steam at 
the beginning of each stroke, and also tends to diminish the con- 
densation of this steam by heating up the cylinder walls, on the 
other hand it has the disadvantage of producing a reversal of 
pressure before the end of each stroke, which tends to produce a 
shock on the crank pin. If we wish to relieve the crank pin of 
all pressure at the end of the stroke we have only to compress 
the exhaust steam till its final pressure is equal to the sum of 
the terminal pressure (near end of the expansion line) and the 
pressure retarding the reciprocating parts. As in high speed 
engines the terminal pressure is practically reduced to a very 
small amount through expansion and an early exhaust, we may 



12 



CONSTRUCTION OF HYPERBOLA. 



call it 1 6 pounds and then we have the rule to compress up to 
the pressure 

F. A + i6 



'' + -A 



approximately, 



(8) 



~ being the mean retarding (or accelerating) pressure of the 
A 

reciprocating parts at the end of the stroke. Stopping the 
compression short of the steam chest pressure has the incidental 
advantage of a good admission line to the indicator card, for 
there exists then a difference of pressure which can hurry the 
steam from the boiler into the cylinder. Accepting the latter 
rule as the one for our guidance we can draw the compression 
curve according to the law pv = constant. When compression 
is carried up to boiler pressure, and cylinder is steam jacketed, 
we should use the law/z/f = constant. 

I D. 

Method of Constructing Equilateral Hyperbola. 
The use of the Mariotte curve or equilateral hyperbola, is so 
common, that an exact method of drawing it may not be out of 
place here, although it is already known to engineers. The 
equation is pv = p^v^ where p^ and v^ are respectively the abso- 
lute pressure and volume of the gas at some point where they 
ar^ known, and p and v those at any other point. From the cen- 
ter O, Fig. I, which is the intersection of the clearance and 

T A 



/ ^ 


X 




<~^ 







£ 

Fig. I. 



D 



CURVE CONSTRUCTION BY TABULATED ORDINATES. 



13 



vacuum lines, whatever the degree of expansion employed, 
draw the line OA and from the points A and C, where it 
cuts the steam and cut-off lines let fall the perpendiculars 
AB and CB. Their intersection at B will give a point of the 
curve exactly. For from the similar triangles AOD and OCE 
we have CE : OE : : AD : OD and consequently BD : OE : : 
FE : OD. But OE and FE represent respectively the volume 
and pressure at the point of cut-off F, and OD and BD those at 
the point B, hence p : v^ .: p^ : v or pv =^ p^v^, therefore the 
point B was correctly found. Other points may be found in the 
same way. 

Method of Determining Points, on the Curves pv^ = 

Constant, by Tabulated Coordinates. 
To find the coordinates for the curves by means of the follow- 
ing table we divide, in Fig. 2, the distance HE ^= AC = v^, into 




J E 



Aim. 



Vac. 



Fig. 2. 



four equal parts and lay off one of these parts as many times as 

HI 



possible on FE. Then will for example 



^/_ 



^i 



NF 



= 1.75, also 



V, FH 



3, etc. At the points of division on FE erect ordinates 



P 
and lay off on them the value of -- X p^ corresponding to the 



14 



ORDINATES OF STEAM CURVES. 



TABLE 11. 

Values of the Ratio- of the Absolute Pressures 

Occurring in Formula p^v^ = pv"^ = Constant. 

for^ in linear measure multiply — by pr expressed in linear 

measure. 



Si 




Exponent n in 


formula /i 2/i « = pv^. 






>^3 d 

I- nS 


S 


h- 




i 


(3 


<o o 

S > 

1 ^ 

V 


lis 

<u L, (u §i 

P 0) (U 

wis 
-1 




11 "^ ^ 

< 


-lis 

II 3 2^ 

^ " 5 

5- 


iig-sl 

M <^ <U _^ 

S ex 
.. 


11 o| 

CO > {fl 
• nT3 

^ ^ 
11.2 g 

-§ 




-3 

II _o 


1. 125 


.8889 


.8822 


.8759 


.8748 


.8631 


.8547 


.8472 


1.25 


.8000 


.7886 


.7780 


■7763 


.7569 


.7427 


.7304 


1.50 


.6667 


.6494 


•6337 


.6312 


.6024 


.5824 


.5650 


1-75 


.5714 


.5511 


.5328 


.5299 


.4968 


.4948 


.4548 


2.00 


.5000 


.4781 


•4585 


.4550 


•4265 


.3969 


.3767 


2.25 


.4444 


.4218 


.4016 


.3984 


.3629 


.3392 


.3192 


2.50 


.4000 


.3771 


•3567 


•3535 


.3121 


.2947 


.2752 


2.75 


.3636 


.3406 


.3204 


.3172 


.2824 


.2595 


.2407 


3.00 


•3333 


•3105 


.2906 


.2874 


.2533 


.2311 


.2129 


3-50 


.2857 


.2635 


•2443 


.2413 


.2089 


.1882 


.1714 


4.00 


.2500 


.2286 


.2102 


.2073 


.1768 


•1575 


.1420 


4.50 


.2222 


.2016 


.1841 


.1814 


.1526 


.1346 


.1203 


5. 


.2000 


.1803 


.1636 


.1548 


•1337 


.1170 


.1037 


6. 


.1657 


.1484 


•1332 


.1309 


.1065 


.0917 


.0802 


7. 


.1429 


.1260 


.1129 


.1099 


.0878 


.0747 


.0646 


8. 


.1250 


.1093 


.0964 


.0944 


.0743 


.0626 


.0535 


9- 


.nil 


.0964 


.0844 


.0826 


.0642 


.0534 


•0453 


10. 


.1000 


.0862 


.0750 


.0733 


.0562 


.0464 


.0391 


II. 


.0909 


.0779 


.0674 


.0658 


.0499 


.0409 


.0342 


12. 


•0833 


.0710 


.0611 


.0596 


.0450 


.0364 


.0302 


13- 


.0769 


.0653 


.0558 


.0544 


.0405 


.0327 


.0270 


14. 


.0714 


.0602 


.0514 


.0500 


.0369 


.0296 


.0243 


15- 


.0767 


.0562 


.0475 


.0463 


.0339 


.0270 


.0221 


16. 


.0625 


.0523 


.0442 


.0430 


•0313 


.0248 


.0206 


18. 


.0556 


.0461 


.0387 


.0376 


.0270 


.0212 


.0175 


20. 


.0500 


.0412 


•0344 


.0334 


.0236 


.0184 


.0147 


22. 


.0455 


.0372 


.0309 


.0300 


.0210 


.0162 


.0129 


24. 


.0417 


.0339 


.0280 


.0271 


.0188 


.0145 


.0114 


26. 


.0385 


.0312 


.0257 


.0248 


.0170 


.0130 


.0102 


28. 


•0357 


.0288 


.0236 


.0228 


•0155 


.0118 


.0092 


30. 


.0333 


.0268 


.0218 


.0211 


.0142 


.0107 


.0083 



DETERMINATION OF MEAN EFFECTIVE PRESSURE. 1 5 

ratio of their abscissas v to z/j. As stated in Table II, if we 
wish to obtain the value of the ordinate p in linear measure 

P 
we must multiply the ratios — by p^ expressed in linear 

A C 
measure. When the cut-off ——-is very small we need not divide 

JrirL 

HF into four equal parts, but can then simply lay off HF as 
many times as possible on FE, thus getting a sufficient number 
of points to draw the curve CD with accuracy. The compres- 
sion curve F^E^, Fig. 2, can be drawn in a similar manner. 

Construct two indicator diagrams, one for each end of the 
cylinder, the initial pressure being assumed equal to 85 pounds 
(by gauge), the real cut-off = ]^ and the clearance = 0.07. 
Leave space on sheet for four more indicator diagrams, two of 
which will represent the maximum, and two the minimum, 
power of the engine. 

I E. 

Determination of Mean Effective Pressure of Indicator 
AND Other Diagrams. 

The problem in this case is simply to find the average height 
of an irregular but closed figure. This may be done in four 
different ways, ist, By John Coffin's modification of the plani- 
meter (see catalogue of Ashcroft Manuf. Co). 2d, By dividing 
the enclosed area by equidistant ordinates and then calculating 
the area by Simpson's rule (see Nystrom, p. 114). 3d, By divid- 
ing the enclosed area by equidistant ordinates and changing the 
area included between each pair of ordinates into an equivalent 
rectangle and then measuring the height of the latter, the mean 
of all these heights will be that required. 4th, When as in in- 
dicator diagrams, the character of the bounding curves is known 
the average height may be calculated by means of the calculus 
and the results tabulated. The third method is most often used 
in "working up" indicator cards (diagrams), though the first is 



i6 



MEAN EFFECTIVE PRESSURE. 



also used when many cards are to be " worked up." Tabulated 
heights (or mean total pressures when clearance, back pressure 
and compression are neglected) are principally useful when lay- 
ing out the indicator diagrams of proposed engines. The largest 
table of the kind with which the writer is acquainted is that pre- 
pared by Mr. Richard Buel for Du Bois' translation of Weis- 
bach's Mechanics, Vol. II, (Heat Engines, etc.), pp. 479 and 482. 
The following Table III was taken from this work. The next 
Table, IV on page 18, was prepared by the writer and does not 
apply to all cases, but only to the case in which the cut-off is 
chosen with special reference to economy of steam. The body of 
the table contains the mean effective pressure p^ of indicator dia- 
grams for different amounts of clearance and real cut-offs £, or 
corresponding initial pressures /i, prescribed by Mr. Emery's for- 
mula. The back pressure was assumed =16 pounds and both 
expansion and compression curves were assumed to follow the 
law pvrk = constant. Compression was supposed to be con- 
tinued till the final pressure = 



A + 16 ^ i^o 



(9) 




Fig. 3. 

//? AS 

— - = real cut-off — -- = apparent cut-off /A = clearance. 
Hjt Gr 

IH^= BK= absolute initial pressure/^, p"^ = IBCFHI ~- HF. 



TOTAL MEAN EFFECTIVE PRESSURES. 



17 



TABLE III. 

Table Containing Values of the Ratio 
FOR THE Steam Curves pv and /z^H. 



/'. 





03 






nj 






03* 






A 






s^ 


^ i 




s-§ 


c' i 




S^ 


c" i 




'0 

4J X) 


s i 


% 

^ 


1^ 

12 >^ 


5 1 







S3 




1 


la 


OS u! 

|3 







o3 Z 

|3 


p 





c 

^ 


3 




2^ 


8§ 





8^ 


°l 


3 



'7;J 


8g 


s 


II _g 


11 'S 


S 


03 
II fe 


H'^ 


"rt 


03 
II 5 


II rt 


S 


2 
II ^ 


IIS 


p^ 


fi ^ 


:^5 


^ 


ti ^ 


"X-^ 


fS 


Ji "c5 




(S 


fi^ 


^Ho 3 






^J? 




^•| 


K^ 






^J^ 








.01 


W 




.26 


w 




•51 


W 




n6 


W 




.0561 


.0500 


.6102 


•5959 


•8534 


.8463 


.9686 


.9668 


.02 


.0982 


.0884 


.27 


.6235 


.6594 


.52 


.8600 


•8532 


■77 


.9713 


.9696 


•03 


.1352 


.1245 


.28 


.6364 


.6226 


•53 


.8665 


•8599 


•78 


•9738 


•9723 


.04 


.1688 


.1566 


.29 


.6490 


•6355 


•54 


•8727 


.8664 


•79 


.9762 


•9749 


.05 


.1998 


.1866 


•30 


.6612 


.6479 


•55 


.8788 


•8727 


.80 


.9785 


•9773 


.06 


.2288 


.2148 


•31 


.6731 


.6601 


.56 


.8847 


•8789 


.81 


.9807 


.9796 


.07 


.2562 


.2415 


•32 


.6846 


.6719 


•57 


.8904 


.8848 


.82 


.9827 


•9817 


.08 


.2821 


.2669 


•33 


.6959 


.6835 


.58 


•8959 


.8906 


.83 


.9847 


.9838 


.09 


.3067 


.2912 


•34 


.7068 


.6947 


.59 


.9013 


.8962 


.84 


.9865 


•9857 


.10 


.3303 


•3145 


•35 


.7174 


.7056 


.60 


.9065 


.9017 


.85 


.9881 


•9874 


.11 


.3528 


.3368 


.36 


.7278 


.7163 


.61 


.9115 


.9069 


.86 


•9897 


.9891 


.12 


.3744 


.3583 


•37 


.7379 


.7267 


.62 


.9164 


.9120 


•87 


.9912 


.9906 


.13 


.3952 


.3790 


.38 


.7477 


.7368 


.63 


.9211 


.9169 


.88 


•9925 


.9920 


.14 


.4153 


•3990 


•39 


.7572 


.7466 


.64 


.9256 


.9217 


.89 


•9937 


.9933 


.15 


•4346 


.4183 


.40 


.7665 


.7562 


.65 


.9300 


.9263 


.90 


•9948 


■9945 


.16 


.4532 


.4370 


.41 


.7756 


.7656 


.66 


•9342 


•9307 


.91 


•9958 


.9956 


M 


.4712 


•4552 


•42 


•7844 


.7747 


.(^1 


•9383 


•9350 


.92 


.9967 


.9965 


.18 


.4887 


.4727 


•43 


.7929 


•7835 


.68 


.9423 


•9391 


•93 


•9975 


.9973 


.19 


.5055 


.4897 


•44 


.8012 


.7921 


.69 


.9460 


•9431 


.94 


.9982 


.9981 


.20 


.5219 


.5062 


•45 


.8093 


.8005 


.70 


•9497 


.9469 


•95 


•9987 


•9987 


.21 


.5377 


•5223 


.46 


.8172 


.8087 


.71 


.9532 


.9506 


.96 


•9992 


.9991 


.22 


.5531 


.5378 


.47 


.8249 


.8166 


.72 


•9565 


•9541 


.97 


.9996 


•9995 


.23 


.5680 


•5530 


.48 


.8323 


.8244 


.73 


•9597 


•9575 


•98 


•9998 


•9998 


.24 


.5825 


.5677 


•49 


•8395 


.8319 


•74 


.9628 


.9607 


•99 


•9999 


•9999 


.25 


.5966 


.5820 


.50 


.8466 


.8392 1 


•75 


.9658 


.9638 









i8 



MEAN EFFECTIVE PRESSURE. 



TABLE IV. 

Values of the Mean Effective Pressures p^n 
Corresponding to Economical Point of Cut-off, 



£ = 



22 



22 + /x 



Compression takes place up to a pressure 



A + i6_ 



Fo 



A-^- 



Here all Steam Curves follow the Law P^v^tq = constant. 



See reference to this Table on page ib. 



Real 














Cut-off 


Absolute 






Clearance. 






I 


pressures 












r 


Pv 


.01 


.03 


.05 


.07 


.09 


.10 


198. 


44.27 


34.31 


26.43 


18.53 


10.65 


.11 


178. 


40.62 


33.94 


27.27 


20.61 


13.96 


.12 


161. 3 


38.93 


33-18 


26.56 


21.67 


15-99 


•13 


147.2 


37.3 


32.16 


27.37 


22.39 


174 


.14 


I35-I 


35.8 


314 


27.1 


22.7 


18.3 


•'1 


124.7 


34.3 


304 


26.6 


22.8 


19.0 


.16 


115.5 


32.8 


29.4 


26.1 


22.7 


19.4 


.17 


107.4 


31.4 


28.4 


25.5 


22.5 


19.5 


.18 


100.2 


30.2 


27.4 


24.8 


22.1 


19.5 


.19 


93.8 


28.8 


26.4 


24,0 


21.7 


19-3 


.20 


88.0 


27.5 


25.4 


23-3 


21.2 


19.1 


.21 


82.8 


26.3 


24.4 


22.5 


20.6 


18.8 


.22 


78.0 


25.1 


23.1 


21.7 


20.0 


18.4 


.23 


73-7 


24.0 


22.5 


20.9 


19.4 


17.9 


.24 


69.7 


22.9 


21.5 


20.0 


18.8 


17.6 


.25 


66.0 


21.8 


20.6 


19.3 


18.1 


16.9 


.26 


62.6 


20.8 


197 


18.5 


174 


16.3 


.27 


59-5 


19.8 


18.7 


17.7 


16.7 


15-7 


.28 


56.6 


18.8 


17.9 


17.0 


16.1 


15. 1 


.29 


53-9 


17.8 


17.0 


16.2 


15.4 


14.6 


.30 


51-3 


16.9 


16.1 


15.4 


15.2 


13-9 



CUT-OFF FROM MEAN EFFECTIVE PRESSURE. 



19 



I F. 

To determine the cut-ofTf corresponding to a given horse-power 
//j, when the mean effective pressure p„i and the mean total pres- 
sure p" ,n (see Table III and Fig. 3) is known for another horse- 
power H of the same engine running at the same speed, with the 
same initial pressure/i and the same back pressure and compression. 

In engines with cut-off under the control of the governor, the 
speed {i.e. R.p.m) remains constant. Hence we have 



TT 

pm '■ pmx '•'■ H : //i and/,;,! = —ip^ 



(10) 



A B 




ON K, JK 



Fig. 4. 



By means of Fig. 4, if we suppose the line of counter-pressures 
DEF to remain the same for all variations of power, we can 
deduce the followincr relations : 



area 



BCDEFB 



stroke HI 

area ACDIGA 

length GI 



= P> 



= P' 
AB 



area BC,LD,MEFB 

stroke HI 
2.x^2.AC^LDJGA 



= pmz (11) 



length GI 



= P"^. (12) 



In the figure, -— - = clearance i and AG =^ initial pressure /j. 



HI 
Table III on page 17 contains the values of ^^— ^' for 



various 



20 CUT-OFF FROM MEAN EFFECTIVE PRESSURE. 

real cut-offs and for the two steam curves pv = constant and 
p^v^ =z constant. 

area ABFEMDIGA 

A =- -^r- — - ^^ 

A = GirvHi =^''--TTi ^'^^ 

p"mr X GI = p,m X HI -{- p^X GI 

,, _ Pm^ I ,„ Pm 

I' mi , • \ F ^ 



i-{-t ^ -\- ^ 




hence^^ = f ^" .f" -f- ^1^ _ il^ -f I ^ \r^. (14) 

In a like manner it may be shown that 




where /"^s is the corresponding mean total pressure for the 
horse-power H^. 

The quantities in the second members of these equations are 
either given directly or may be found from the diagrams already 
constructed or from the tables. This enables us to calculate 






P"mi. P' Tn^ 

^—^ 2Xidi^—^ and from these, by means of Table III on page 
/i A 

17, we can get the corresponding cut-offs — and — • 



Calculate the cut-offs corresponding to 50 and 170 I.H.P., the 
extremes of power for the proposed engine. Then draw two 
indicator diagrams for each of these powers. 



DEFINITION OF FORWARD AND RETURN STROKES. 



21 



I G. 

Diagrams of Effective Steam Pressures. 
Draw diagram of effective steam pressures on piston both for 
the forward and return strokes. We will here state what is to 
be understood by the forward and the return stroke. In all cases 
suppose the engine to lie horizontally, with the crank on the right 
hand and the cylinder on the left. The forward stroke will then 
be when the piston moves to the right or when the crank 
moves in its upper or lower semi-circle so that it first passes 
through the quadrant nearest the cylinder and then through the 
quadrant farthest from the cyHnder. This does not agree with 
locomotive practice where the forward end of the cylinder is its 
front end, and the stroke of the piston toward that end the for- 
ward stroke, but it agrees with the practice of many engineers 
when treating of stationary engines. To avoid all ambiguity as 



,Jach £nd 



fhiui-afd. 



n 



xvvwvvwvv 







Fig. 5. 



to which end of the cylinder is meant, it would be well when fig- 
ures are not given to speak of the end in question as nearest to, 
or farthest from, the crank. In like manner we may speak of 
the stroke toward or away from the crank shaft, which will cor- 
respond respectively to forward and return strokes as defined 
above. 



22 



DIAGRAMS FROM BACK AND FRONT ENDS. 



Fig. 6 and Fig. 7 represent respectively the indicator diagrams 
from back and front ends of the cylinder shown in Fig. 5. If 
we wish to ascertain only the power of an engine these indica- 
tor diagrams will be sufficient, for from them we can obtain in 
the usual manner, the power developed by the steam in each 
end of the cylinder, but if we wish to ascertain the real pressure 




exerted by the steam upon the piston at any position of the 
latter we must combine both diagrams as follows. 

The admission and expansion hne of one end of the cyHnder 
being placed upon the exhaust and compression line of the other 
end, Figs. 8 and 9, the portions of the vertical ordinates included 



EFFECTIVE STEAM PRESSURE. 



23 



by the shaded portion of these figures will represent the difference 
between the absolute pressures of the steam on opposite sides of 
the piston and consequently will represent the effective pressure 
or driving effort exerted by the steam upon the piston for the 
various positions of the latter. For instance, for the piston posi- 
tion in Fig. 5, on the forward stroke, the effective or driving 




Tonc-ard ptroke 



Fig. 8. 




Fig. 9. 

pressure on the piston will be M^N' = M'O' — N^0\ Fig. 8, 
M^O^ being the absolute pressure of steam in the back end of 
cylinder, and N'O^ the absolute pressure in the front end. The 
intercept M^^N'\ Fig. 6 or 7, corresponding to the same piston 
position is considerably different from M^N\ Fig. 8 or 9, and does 



24 RATIO OF ROD TO CRANK. 

not represent the difference of two pressures existing simulta- 
neously, in opposite ends of the cylinder, but represents the differ- 
ence between two pressures separated by an interval of time cor- 
responding to about half a revolution of the crank. The effec- 
tive driving pressure on the piston for the position in Fig. 5, 
on the return stroke is equal to zero, as is shown at M'^N^^ 
Fig. 9, the pressure on opposite sides of piston being equal at that 
moment. The point where the lines cross represents, therefore 
equilibrium between the opposing pressures ; after this point is 
passed the resistances are in excess, and if it were not for the in- 
ertia of the reciprocating parts and fly-wheel, the motion of the 
piston would be reversed before it had completed its stroke. This 
change in the direction of the excess of pressure is represented 
in the diagram by a difference in shading. This reversal of pres- 
sures serves a most useful purpose in absorbing the inertia of the 
reciprocating parts, bringing them gradually to rest and thus 
preventing shocks and vibrations at the end of the stroke. Draw 
in accordance with the above, three pairs of diagrams of effective 
steam pressures on piston, corresponding to 50, no and 170 
I.H.P., the minimum, normal and maximum horse-powers of this 
engine. 

II. 

Determination of Ratio of Length of Connecting Rod 
TO Length of Crank. 

When the connecting rod is infinite in length (which is the 
case in the mechanism known as the normal double slider crank 
chain, Reuleaux's Kinematics, p. 314) the speed of the piston is 
the same for the crank positions co and 180° — co ; the distance 
from the nearest end of the stroke is also the same for these two 
crank positions, for the piston will travel through the first and 
second halves of its stroke while the crank pin is sweeping 
through the first and second quadrants of its motion. The hori- 
zontal motion of the piston along the axis of the cylinder corres- 



INFLUENCE OF ROD ON VELOCITY. 



25 



ponds exactly to the horizontal motion of the crank pin at 
every instant. Moreover, the pressure on slides at right angles 
to piston motion = o, the obliquity of the connecting rod 
being o. 

But when the connecting rod is of finite length, the piston 
speed and distances from nearest end of stroke are different 
for the crank angles co and 1 80° — co, the paths traversed by piston 
while the crank is swinging through the first and second quad- 
rants are unequal, and the pressure on the slide and conse- 
quently the friction on the slide is proportional to the ratio of 
the length of crank to connecting rod. In the following figure 




Fig, 10. 



we have taken 



connecting rod 



= 2 ^ to show the effect of a very 



crank 
short rod on piston speeds and positions. 

If we supposse the motion of the crank pin uniform and equal 
to V we shall have the piston speed w =^ v sin co for each of the 
crank positions S and T when the connecting rod is infinite in 
length, but when it is finite and equal to 2}4 times the crank, as 
in the figure, the velocity of piston for crank positions S and T 
will be respectively (see Fig. 10 for co) 



V sm 0) 



{ 1 H ^ — -J for point T 

A ' f COS co\ 

and w = 7J sm co I i J for pomt 5. 



(16) 



26 



HIGH-SPEED STEAM ENGINE. 



TABLE V. 



Distance of Piston from Beginning of Stroke when Stroke is 
towards Crank-shaft. 

To find Actual Distance of Piston from end of Stroke, multiply tabular quantity 

by Stroke. 



•^i 


Connecting-rod ^- Crank = 


-^J 




Connecting-rod -h- Crank 


= 




2^ 






















4 


4.5 


5 


5.5 


6 7 8 


00 


4 


4.5 


5 


5.5 


6 


7 


8 


00 


2 


.0004 


.0004 


.0004 


.0004 


.0004 .0003 .0003 


.0003 


92 


.5809 


.5737 


.5679 


.5632 


.5594 


.5533 


.5487 


5174 


4 


.0015 


.0015 


.0015 


.0014 


.0014 .0014 .0014 


.0012 


94 


.5981 


.5909 


.5851 


.5805 


.5766 


.6706 


.5661 


.5349 


6 


.0034 


.0033 


.0033 


.0032 


.0032 


.0031 .0030 


.0027 


96 


.6151 


.6079 


.6022 


.5977 


.5938 


.5878 


.5833 


.5523 


8 


.0061 


.0060 


.0058 


.0057 


.0057 


.0056 .0055 


.0049 


98 


.6318 


.6247 


.6191 


.6145 


.6107 


.6048 


.6004 


.5696 


10 


.0095 


.0092 


.0091 


.0089 


.0089 


.00871.0085 


.0076 


100 


.6484 


.6414 


.6358 


.6313 


.6275 


.6216 


.6172 


.5868 


12 


.0136 


.0133 


.0131 


.0129 


.0127 


.0126 .0124 


.0109 


102 


.6647 


.6578 


.6523 


.6478 


.6441 


.6384 


.6340 


.6040 


14 


.0185 


.0181 


.0178 


.0175 


.0173 


.0170 .0167 


•0149 


104 


.6807 


.6739 


.6685 


.6641 


.6605 


.6548 


.6505 


.6210 


16 


.0241 


.0236 


.0232 


.0228 


.0225 


.02211.0218 


.0194 


106 


.6964 


.6897 


.6845 


.6802 


.6766 


.6710 


.6668 


.6378 


18 


.0304 


.0298 


.0293 


.0290 


.0285 


.0279 .0275 


.0245 


108 


.7119 


.7053 


.7002 


.6959 


.6924 


.6870 


.6829 


.6545 


20 


.0375 


.0367 


.0360 


.0355 


.0350 


.0344 .0339 


.0302 


110 


.7270 


.7206 


.7115 


.7115 


.7080 


.7027 


.6987 


.6710 


22 


.0452 


.0442 


.0434 


.0428 


.0423 


.0414 .0408 


.0364 


112 


.7418 


.7356 


.7307 


.7267 


.7233 


.7181 


.7143 


.6873 


24 


.0536 


.0524 


.0515 


.0507 


.0501 


.0491 .0484 


.0432 


114 


.7562 


.7502 


.7455 


.7416 


.7383 


.7334 


.7296 


.7034 


26 


.0627 


.0613 


.0602 


.0594 


.0586 


.0575 .0566 


.0506 


116 


.7703 


.7645 


.7599 


.7561 


.7530 


.7482 


.7445 


.7192 


28 


.0724 


.0708 


.0696 


.0686 


.0677 


.0664 .0654 


.0585 


118 


.7841 


.7785 


.7740 


.7704 


.7674 


.7626 


.7591 


.7347 


30 


.0827 


.0809 


.0796 


.0784 


.0774 


.0759 .0748 


.0670 


120 


.7974 


.7921 


.7878 


.7843 


.7814 


.7769 


.7735 


.7500 


32 


.0936 


.0916 


.0901 


.0888 


.0877 


.0860 .0848 


.0760 


122 


.8104 


.8053 


.8012 


.7979 


.7951 


.7908 


.7875 


.7650 


34 


.1051 


.1029 


.1012 


.0997 


.0985 


.09671.0953 


.0855 


124 


.8230 


.8181 


.8142 


.8105 


.8084 


.8042 


.8011 


.7796 


36 


.1172 


.1147 


.1128 


.1112 


.1099 


.1079|.1063 


.0955 


126 


.8352 


.8305 


.8268 


.8238 


.8213 


.8173 


.8144 


.7939 


38 


.1298 


.1271 


.1250 


.1233 


.1218 


.1196, .1179 


.1060 


128 


.8470 


.8426 


.8361 
.8509 


.8362 


.8338 


.8300 


.8273 


.8078 


40 


.1430 


.1400 


.1317 


.1358 


.1342 


.1318|.1297 


.1170 


130 


.8584 


.8542 


.8452 


.8459 


.8424 


.8398 


.8214 


42 


.1566 


.1534 


.1509 


.1489 


.1471 


.1444'. 1424 


.1284 


132 


.8694 


.8655 


.8623 


.8598 


.8577 


.8544 


.8519 


.8346 


44 


.1707 


.1673 


.1646 


.1623 


.1605 


.1576'. 1557 


.1403 


134 


.8799 


.8763 


.8733 


.8710 


.8690 


.8658 


.8635 


.8473 


46 


.1853 


.1816 


.1787 


.1763 


.1743 


.1712 .1689 


.1527 


136 


.8901 


,8866 


.8839 


.8817 


.8798 


.8770 


.8751 


.8597 


48 


.2003 


.1963 


.1922 


.1905 


.1885 


.1852 .1827 


.1654 


138 


.8998 


.8966 


.8941 


.8920 


.8903 


.8876 


.8856 


.8716 


50 


.2156 


.2115 


.2081 


.2054 


.2032 


.19961.2970 


.1786 


140 


.9090 


.9061 


.9038 


9019 


.9003 


.8978 


.8959 


.8830 


52 


.2314 


.2270 


.2234 


.2206 


.2182 


.2144 .2117 


.1922 


142 


.9178 


.9152 


.9130 


.9113 


.9098 


.9076 


.9059 


.8940 


54 


.2474 


.2428 


.2391 


.2360 


.2335 


.2293 .2266 


.2061 


144 


.9262 


.9238 


.9218 


.9203 


.9189 


.9169 


.9153 


.9045 


56 


.2638 


.2589 


.2550 


.2518 


.2492 


.2450 .2419 


.2204 


146 


.9342 


.9320 


.9302 


.9288 


.9276 


.9257 


.9243 


.9145 


58 


.2805 


.2754 


.2713 


.2679 


.2652 


.2608;. 2575 


.2350 


148 


.9417 


.9397 


.9381 


.9368 


.9358 


.9340 


.9328 


.9240 


60 


.2974 


.2920 


.2878 


.2843 


.2814 


.2769 .2735 


.2500 


150 


.9487 


.9469 


.9455 


.9444 


.9435 


.9419 


.9408 


.9330 


62 


.3146 


.3090 


.3046 


.3009 


.2979 


.2932 .2897 


.2653 


152 


.9553 


.9538 


.9825 


.9515 


.9507 


.9494 


.9484 


.9415 


64 


.3320 


.3262 


.3215 


.3178 


.3147 


.30981.3061 


.2808 


154 


.9617 


.9601 


.9590 


.9581 


.9574 


.9563 


.9554 


9494 


66 


.3495 


.3435 


.3387 


.3348 


.3316 


.3266'. 3228 


.2966 


156 


.9671 


.9660 


.9651 


.9643 


.9637 


.9627 


.9620 


.9568 


68 


.3672 


.3610 


.3561 


.3520 


.3487 


.3435; .3397 


.3127 


158 


.9724 


.9713 


.9706 


.9700 


.9694 


.9686 


.9678 


.9634 


70 


.3850 


.3786 


.3735 


.3694 


.3660 


.3607 .3567 


.3290 


160 


.9772 


.9763 


.9757 


.9751 


.9747 


.9740 


.9735 


.9698 


72 


.4028 


.3963 


.3911 


.3869 


.3834 


.3780. 3739 


.3455 


162 


.9815 


.9808 


.9803 


.9801 


.9795 


.9789 


.9785 


.9755 


74 


.4208 


.4141 


.4088 


.4045 


.4009 


.3954 .3912 


.3622 


164 


.9854 


.9848 


.9844 


.9841 


.9838 


.9833 


.9830 


.9806 


76 


.4388 


.4319 


.4266 


.4222 


.4185 .4128,. 4085 


.3790 


166 


.9888 


.9884 


.9881 


.9878 


.9878 


.9872 


.9869 


.9851 


78 


.4568 


.4498 


.4444 


.4399 


.4362 .4304 .4260 


.3960 


168 


.9918 


.9913 


.9912 


.9910 


.9909 


.9907 


.9905 


.9891 


80 


.4747 


.4677 


.4622 


.4576 


.4539 .4480 .4436 


.4132 


170 


.9943 


.9941 


.9939 


.9938 


.9937 


.9935 


.9933 


.9924 


82 


.4927 


.4857 


.4799 


.4754 


.4716 .4656!. 4612 


.4304 


172 


.9964 


.9962 


.9961 


.9960 


.9959 


.9958 


.9957 


.9951 


84 


.5105 


.5034 


.4977 


.4930 


.4892 .4832 .4787 


.4477 


174 


.9979 


.9979 


.9978 


.9978 


.9977 


.9977 


.9976 


.9973 


86 


.5283 


.5211 


.5154 


.5107 


.5069 .5008*. 4963 


.4651 


176 


.9990 


.9990 


.9990 


.9990 


.9990 


.9990 


.9990 


.9988 


88 


.5460 


.5388 


.5330 


.52831.5245 .5185 .5139 


.4826 


178 


.9998 


.9997 


.9997 


.9997 


.9997 


.9997 


.9997 


.9997 


90 


.5635 


.5563 


.5505 


.54531.5420 .5359 .5314 


.5000 


180 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 



HIGH-SPEED STEAM ENGINE. 



27 



TABLE VI. 



Distance of Piston from Beginning of Stroke when Stroke is 
away from Crank-shaft. 

To find Actual Distance of Piston from end of Stroke, multiply tabular quantity 

by Stroke. 



•SJ 


Connecting-rod -^ Crank = 


-^i 




Connecting-rod -f- Crank 


=: 




51 

2 






















4 


4.5 


5 


5.5 


6 


7 
.0003 


8 


00 


4 


4.5 


5 


5.5 


6 


7 


8 


00 


.0002 


.0003 


.0003 


.0003 


.0003 


.0003 


.0003 


92 


.4540 


.4612 


.4670 


.4717 


.4755 


.4715 


.4861 


.5174 


4 


.0010 


.0010 


.0010 


.0010 


.0010 


.0010 


.0010 


.0012 


94 


.4717 


.4789 


.4846 


.4893 


.4931 


.4992 


.5037 


.5349 


6 


.0021 


.0021 


.0022 


.0022 


.0023 


.0023 


.0024 


.0027 


96 


.4895 


.4966 


.5023 


.5070 


.5108 


.5168 


.5213 


.5523 


8 


.0036 


.0038 


.0039 


.0040 


.0041 


.0042 


.0043 


.0049 


98 


.5073 


.5143 


.5201 


.5246 


.5284 


.5344 


.5388 


.5696 


10 


.0057 


.0059 


.0061 


.0062 


.0063 


.0065 


.0067 


.0076 


100 


.5253 


.5323 


.5378 


.5424 


.5461 


.5520 


.5564 


.5868 


12 


.0082 


.0087 


.0088 


.0090 


.0091 


.0093 


.0095 


.0109 


102 


.5432 


.5502 


.5556 


.5601 


.5638 


.5696 


.5740 


.6140 


14 


.0112 


.0116 


.0119 


.0122 


.0122 


.0128 


.0131 


.0149 


104 


.5612 


.5681 


.5734 


.5778 


.5815 


.5872 


.5915 


.6210 


16 


.0146 


.0152 


.0156 


.0159 


.0162 


.0167 


.0170 


.0194 


106 


.5792 


.5859 


.5912 


.5955 


.5991 


.6046 


.6088 


.6378 


18 


.0185 


.0192 


.0197 


.0199 


.0205 


.0211 


.0215 


.0245 


108 


.5972 


.6037 


.6089 


.6131 


.6166 


.6220 


.6261 


.6545 


20 


.0228 


.0237 


.0243 


.0248 


.0253 


.0260 


.0265 


.0302 


110 


.6150 


.6214 


.6265 


.6306 


.6340 


.6393 


.6533 


.6710 


22 


.0276 


.0286 


.0294 


.0300 


.0306 


.0314 


.0322 


.0366 


112 


.6328 


.6390 


.6439 


.6480 


.6513 


.6565 


.6603 


.6873 


24 


.0329 


.0340 


.0349 


.0367 


.0363 


.0373 


.0380 


.04321 


114 


.6505 


.6565 


.6613 


.6652 


.6684 


.6734 


.6772 


.7034 


26 


.0385 


.0399 


.0410 


.0419 


.0426 


.0437 


.0446 


.0506 


116 


.6680 


.6738 


.6785 


.6822 


.6853 


.6902 


.6939 


.7192 


28 


.0447 


.0462 


.0475 


.0485 


.0493 


.0506 


.0516 


.0585 


118 


.6844 


.6910 


.6954 


.6991 


.7021 


.7068 


.7103 


.7347 


30 


.0513 


.0531 


.0545 


.0556 


.0565 


.0581 


.0592 


.0670! 


120 


.7026 


.7080 


.7122 


.7157 


.7186 


.7231 


.7265 


.7500 


32 


.0583 


.0603 


.0619 


.0632 


.0642 


.0660 


.0671 


.0760' 


122 


.7195 


.7245 


.7287 


.7321 


.7348 


.7392 


.7425 


.7650 


34 


.0658 


.0680 


.0698 


.0712 


.0724 


.0748 


.0757 


.0855 


124 


.7362 


.7411 


.7450 


.7482 


.7508 


.7550 


.7581 


.7796 


36 


.0738 


.0762 


.0782 


.0797 


.0811 


.0831 


.0847 


.0955: 


126 


.7526 


.7572 


.7609 


.7640 


.7665 


.7705 


.7734 


.7939 


38 


.0822 


.0848 


.0870 


.0887 


.0902 


.0924 


.0941 


.1060' 


128 


.7686 


,7730 


.7766 


.7794 


.7818 


.7856 


.7883 


.8078 


40 


.0910 


.0939 


.0962 


.0981 


.0997 


.1022 


.1041 


.1170 


130 


.7844 


.7885 


.7919 


.7919 


.7969 


.8004 


.8030 


.8214 


42 


.1002 


.1034 


.1059 


.1080 


.1097 


.1124 


.1144 


.1284 


132 


.7997 


.8037 


.8068 


.8094 


.8115 


.8148 


.8173 


.8346 


44 


.1099 


.1134 


.1161 


.1183 


.1202 


.1230 


.1249 


.1403 


134 


.8147 


.8184 


.8213 


.8237 


.8257 


.8288 


.8311 


.8473 


46 


.1201 


.1237 


.1267 


.1290 


.1310 


.1342 


.1365 


.1527 


136 


.8293 


.8327 


.8354 


.8377 


.8395 


.8424 


.8443 


.8597 


48 


.1306 


.1348 


.1377 


.1402 


.1423 


.1456 


.1481 


.1654 


138 


.8434 


.8469 


.8491 


.8511 


.8529 


.8556 


.8576 


.8716 


50 


.1416 


.1458 


.1491 


.1518 


.1541 


.1576 


.1607 


.1786 


140 


.8570 


.8600 


.8623 


.8642 


.8658 


.8682 


.8703 


.8830 


52 


.1530 


.1574 


.1609 


.1638 


.1662 


.1700 


.1727 


.1922 


142 


.8702 


.8729 


.8750 


.8767 


.8782 


.8804 


.8821 


.8940 


54 


.1648 


.1695 


.1732 


.1762 


.1787 


.1827 


.1856 


.2061 


144 


.8828 


.8853 


.8882 


.8888 


8901 


.8921 


.8937 


.9045 


56 


.1769 


.1819 


.1858 


.1804 


.1916 


.1958 


.1989 


.2204, 


146 


.8949 


.8971 


.8988 


.9003 


.9015 


.9033 


.9047 


.9145 


58 


.1896 


.1947 


.1988 


.2021 


.2049 


.2092 


.2125 


.2350) 


148 


.9064 


.9084 


.9099 


.9112 


.9123 


.9140 


.9152 


.9240 


60 


.2026 


.2079 


.2122 


.2157 


.2186 


.2231 


.2295 


.2500 


150 


.9173 


.9191 


.9204 


.9216 


.9226 


.9241 


.9252 


.9330 


62 


.2159 


.2215 


.2260 


.2296 


.2326 


.2324 


.2409 


.2653 


i 152 


.9276 


.9292 


.9304 


.9314 


.9323 


.9336 


.9346 


.9415 


64 


.2297 


.2355 


.2401 


.2439 


.2470 


.2518 


.2555 


.2808 


i 154 


.9373 


.9387 


.9398 


.9406 


.9414 


.9425 


.9434 


.9494 


66 


.2438 


.2498 


.2545 


.2584 


.2617 


.2666 


.1704 


.2966 


156 


.9464 


.9476 


.9485 


.9493 


.9499 


.9509 


.9516 


.9568 


68 


.2582 


.2644 


.2693 


.2733 


.2767 


.2819 


.2857 


.3127 


158 


.9548 


.9558 


.9566 


.9572 


.9577 


.9586 


.9592 


.9636 


70 


.2730 


.2794 


.2844 


.2885 


.2929 


.2973 


.3013 


.3290 


160 


.9625 


.9633 


.9640 


.9645 


.9650 


.9656 


.9661 


.9698 


72 


.2881 


.2947 


.2998 


.3041 


.3076 


.3130 


.3171 


.3455 


162 


.9696 


.9752 


.9707 


.9710 


.9715 


.9721 


.9725 


.9755 


74 


.3036 


.3103 


.3155 


.3198 


.3234 


.3290 


.3332 


.3622 


164 


.9759 


.9764 


.9767 


.9772 


.9775 


.9779 


.9782 


.9806 


76 


.3193 


.3261 


.3315 


.3359 


.3395 


.3452 


.3495 


.3790 


. 166 


.9815 


.9819 


.9822 


.9825 


.9827 


.9830 


.9883 


.9851 


78 


.3353 


.3422 


.3477 


.3522 


.3559 


.3616 


.3660 


.3960 


168 


.9864 


.9867 


.9869 


.9871 


.9873 


.9874 


.9876 


.9861 


80 


.3516 


.3586 


.3642 


.3687 


.3725 


.3784 


.3828 


.4132 


j 170 


.9905 


.9908 


.9909 


.9911 


.9912 


.9913 


.9915 


.9924 


82 


.3682 


.3753 


.3809 


.3855 


.3893 


'.3952 


.3996 


.4304 


172 


.9939 


.9940 


.9942 


.9943 


.9943 


.9944 


.9945 


.9951 


84 


|.3849 


.3921 


.3978 


;.4024 


.4062 


.4122 


.4167 


.4477 


174 


.9966 


.9967 


.9967 


.9968 


.9968 


.9969 


.9970 


.9973 


86 


1.4019 


.4091 


.4149 


.4195 


.4234 


.4294 


.4349 


.4651 


1 176 


.9985 


.9985 


.9985 


.9986 


.9986 


.9986 


.9986 


.9988 


88 


1.4191 


.4263 


.4321 


i.4368 


.4406 


:.4467 


,.4013 


.4826 


178 


, .9996 


, .9996 


.9996 


.9996 


.9996 


.9997 


.9997 


.9997 


90 


1.4365 


1.4437 


1.4495 


1.4547 


1.4580 


1.4641 


1.4686 


.5000 


1 180 


1.0000 


I. 0000 


1.0000 


1 1.0000 


1.0000 


I. 0000 


I. 0000 


1.0000 



28 



INFLUENCE OF ROD ON PISTON TRAVEL. 



With respect to distance of piston from end of stroke it is evi- 
dent from the figure that the distances mB and nF, correspond- 
ing to the crank positions 2" and 5, are unequal. If the connect- 
ing rod were infinite rn^B'^ and n^F'^ would respectively represent 
the distances of piston from nearest end of stroke for the crank 
positions Zand vS. 

The following figure, drawn to scale, shows graphically the 
inequalities of the paths traversed by piston during the first and 



d 


S 


M 







h 




i 




'A 








CL 


7 


M 


'A 







Ir 




I 




^ 








a 


6 


M 


^ 







h 




/ 




^ 








a. 


6- 


M 


% 


\o 




I- 




/ 




'/' 








(L 


y 


J^ 


^ 


\o 




h 




/ 




V 








CL 


3 


J^ 


''/. 


/\o 




h 




/ 




^ 


\ 






(X. 


«8 


7^ 


V 


///\o 




Ir 




/ 




1 


vy^ 






a 


/ 


J^ 


1 


'/m 




2=:===™^ 



Fig. II. 

second quadrants of crank motion for different ratios of connect- 
ing rod to crank, the horizontal position of stroke (= ab) included 
by shaded figure representing the distance Mo of the piston from 
its middle position M{sgq preceding figure and Tables V and VI). 

When length of crank =^ R = i and (o = 90° 

L _ 



Mo. 



.063 



1 7 


6 


5 


4 i 3 


2 
.268 


Ion 


.084 


.101 


•127 1. 172 



I.O 



Fig. 1 1 shows clearly that as the ratio of connecting rod to 
crank diminishes the inequality of the paths ao and ob respec- 



INFLUENCE OF ROD ON ROTATIVE EFFECT. 



29 



tively described by piston while crank passes through the first 
and second quadrant, increases. This inequality tends to heap 
up the work in the first and fourth quadrants, as will now be 
more fully shown. 

In order to show how the length of connecting rod influences 
the tangential components of the pressures on crank pin, we will 
assume the horizontal pressure on cross-head pin constant, and 







/> 


y^ 


\ 


/ 


\ 




/j^ 




\ 


/ 


^N^ \ 




/x ^ 




\ 




/^^\ 




l^ ^ 




\ 


/ 


/ >\ 


^7 


r 


\ 




/ 


/ \. 


V 


\ 


^ 




/ \\ 


/ 


X 


1 


^^^ 




{ '' \ 




/ 






V' \ 


-^ / 


/ 








\ / \ 


. / 


L / 








\ --/ \ 


^ 




v"^ 7 








\ / \ 


c 




\ / 








\ / 


<i 




\ / 








/ 


*Ob 












/ 


t 


_ . — 


/t~ " 


'"""' 


— - 






^ 


i 


/\ 








\ 


«o 


\ 


/''\ 








A 0\ / 


>\ 


f \ 








/ \ / 


°^vi 




\ 






cV 


\ 


\ 


/ 


/~ 


"^ 


\ y 




V / 




/ 


\ 


\ A 




\V 




1 


\ 


\/l 




\Av 




1 


\ 


^y\/ 




\ 


"^"v^ 


«/ 




^^ y 






\ 




/"" 


-^ 


y 



Fig. la. 

will lay off the tangential pressures on prolongation of the radii 
of the crank-pin circle, Figs. 12 and 13. The curve drawn through 
the extremities of these radial ordinates will show the variation of 
the tangential pressure for the different crank positions ; the area 
included between this curve and the crank-pin circle measures, ap- 
proximately, the work done in one revolution. The two diagrams 

r Yc\c\ 

infinity and 



, 1 .• 1 . connectmg rod 

drawn correspond respectively to ^ 

crank 

connectins: rod w u a 1 *.4.u j- i, 

= 2 )4 as above. A glance at these diagrams shows 



crank 



30 



INFLUENCE OF ROD ON ROTATIVE EFFECT. 



at once that while with the infinite rod the work done in ^ach of 
the quadrants is equal, with the finite rod the principal part of the 
work is done in quadrants I and IV. The average tangential 
pressures can easily be found by remembering that if friction is 
neglected the work done on piston must equal the work done on 
crank pin, and that therefore the average tangential pressure pt 




Fig. 13. 



must be to the average piston pressure p^ as the path 25 

(S = stroke) of piston in one revolution is to the path nS of the 

crank pin, or 

2 
pi : pm '■: 2S : TiS hence p^ = - p^ = 0.6366 p^. (17) 



Attention has already been called to the fact that the pressure 
and friction on the slide increase with the ratio of crank to con- 
necting rod, then the 



INFLUENCE OF ROD ON STEAM DISTRIBUTION. 3 1 

cr3.nk 

max. friction on slides = aP X ; (i8) 

connecting rod 

^ = coefficient of friction, P = pressure on cross-head pin. 

But the ratio of crank to connecting rod affects not only the 
distribution of work and the magnitude of friction, it also affects 
the distribution of steam by increasing the duration of admis- 
sion and reducing that of exhaust at one end of the cylinder, 
while it diminishes the period of admission and increases the 
duration of exhaust at other end. 

That it thus affects the admission of steam can be roughly 
shown by means of Fig. lo, page 25. Since the ratio of eccen- 
tric radius to length of eccentric rod is small, even when the ratio 

crank 

-. is s^reat, we may neglect the obliquity of eccentric 

connectmg rod ^ ' ^ ^ ^ ^ 

rod and treat the latter as if it were of infinite length, conse- 
quently the angular positions of the eccentric radius for the points 
of cut-off will be diametrically opposite, and since eccentric and 
crank are keyed to the same shaft, the crank positions corre- 
sponding to points of cut-off will also be directly opposite. Let 
//and T, Fig. 10, page 25, represent the two crank positions 
corresponding to cut-off, it will then be evident from the figure 
that the period of admission Bm for the end of the cylinder 
farthest from the crank will be greater than the admission Fn for 
end of cylinder nearest crank. When the ratio of crank to con- 
necting rod is large this distribution of steam will be injurious, 
but when it is equal to, or less than 1 this irregularity of admis- 
sion and exhaust at the two ends will not only not be excessive 
but will be a positive advantage in high-speed engines, because 
it gives the greater compression {i.^. diminished period of exhaust) 
at the end farthest from the crank where it is most needed. This 
will be proved later when we come to consider the proper period 
of compression. Moreover, with such proportions of crank to 
connecting rod, the valve setting obtained from Zeuner's simple 
valve diagram will give very satisfactor}^ results practically. The 



32 RATIO OF ROD TO CRANK. 

indicator diagrams for the two ends of the cyhnder obtained 
from such distribution of steam will of course not be identical 
in form, but they will be nearly equal in area, and as we have 
already mentioned will give the greater compression where it is 
most needed. In marine engines where engine space is limited 
the ratio is sometimes as low as i^, while in other well designed 
stationary engines it is sometimes as high as 8. Decide upon 
the proper ratio for the present case. 



III. 

Determination of that Mean Accelerating Force (per □'' 

OF Piston) Necessary to Start Reciprocating Parts, 

WHICH Corresponds to the Most Uniform 

Tangential Pressure on Crank Pin. 

That the weight and speed of the reciprocating parts have an 
important influence on the steadiness with which an engine can 
be run, will be evident from the following consideration : Steadi- 
ness of running, other things being equal, results from the 
driving effort, or tangential pressure on crank pin, being as 
nearly constant as possible for all positions of the crank ; the 
magnitude of the tangential pressure (= DF in Fig. 14) depends 
upon the position of the crank CD, and upon the direction and 
intensity of the pressure DB' = EB exerted by the connecting 
rod upon the crank pin ; this latter pressure in turn depends upon 
the angular position of the connecting rod and upon the magni- 
tude of the force K =:= EA. Let us now suppose the mass ^of 
the reciprocating parts (piston, piston rod, cross-head and con- 
necting rod)* to be concentrated in a ring around the cross-head 

* According to Weisbach-Herrmann, (Vol. Ill, Section II, p. 867) the 
work stored up in a connecting rod of uniform section is the same as if one- 
third of its mass were concentrated at crank pin and the remaining two- 
thirds at cross-head pin. This result is itself an approximation arid as sec- 
tions are generally not uniform the accuracy of the assumption is still further 
diminished. Usually rod tapers from wrist to crank pin. 



HORIZONTAL PRESSURE ON CRANK PIN. 



33 



pin, and that the resistance offered by this mass to a change in its 
state of motion, is represented by F, we will then have F:= Md 
where 6 represents the acceleration of the center of the cross- 
head pin, M ^= mass of reciprocating parts and F =^ force neces- 
sary to give the mass ^the acceleration 6. 




Fig. 14 



The accelerating force F is of course part of the steam pressure 
P, the difference between them, P — F =z K ^= EA (see figure), 
is that part of the steam pressure which is transmitted through 
the piston rod to the lower end of connecting rod and from these 
to the crank pin. When the ^omtE{i.e., the piston) has reached 
its maximum velocity, acceleration ceases, and we have /^= 
and P ■= K = EA = DA', but retardation now begins (for the 
reciprocating parts must be brought to rest at the end of the 
stroke) and no part of the steam pressure being absorbed in pro- 
ducing acceleration, the whole of the steam pressure Pis trans- 
mitted to the cross-head pin. This P is however not the only 
pressure to which the crank pin is subjected, for the reciprocating 
parts having once attained their maximum velocity, tend to keep 
it, and tend to move faster (horizontally) than the crank pin, and 
will thus exert a pressure Pupon the latter, over and above that 
received from the steam [= P) hence 

P-f- F=^ K= EA = DA\ 



34 HORIZONTAL PRESSURE ON CRANK PIN. 

The general expression for the horizontal pressure exerted 
upon the crank pin or cross-head pin will then be 

DA^ = EA = K=P±F 

the upper sign of F being employed when reciprocating parts are 
retarded, and the lower sign, when the reciprocating parts are 
accelerated, F being the accelerating or retarding force necessary to 
produce in the mass M the acceleration or retardation <9, that is, 

W 
F=Md=-d (19) 

so that F increases with the weight W of the reciprocating parts 
and their acceleration. 

Before determining the value of for the different piston and 
crank positions, we will call attention to, and emphasize the 
point, that the horizontal pressures on crank pin or cross-head 
pin are not directly obtainable from the diagram of effective steam 
pressures on piston but can be obtained indirectly from the latter 
hy subtracting the. accelerating pressures from the steam pressures 
given by the diagram /i?/' the earlier part of the stroke, and adding 
the retardi^ig pressures to the steam pressures given by the dia- 
gram for the later part of the stroke. From this we already sec 
that the mass and motion of the reciprocating parts tend to 
equalize the horizontal pressures upon the crank pin by diminish- 
ing the larger pressures near the beginning and increasing the 
smaller pressures near the end of expansion, the energy of the 
steam absorbed by the reciprocating parts near the beginning, 
being given out again to the crank pin toward the end, of the 
expansion. With too great a speed, however, this equalizing 
tendency will be more than neutralized by the transfer of the 
whole of the steam pressures to the end of the stroke and will 
tend to cause shocks and dangerous stresses on all the working 
pieces. This point will be more fully described further on. 

We will now find an expression for the acceleration Q and then 

substitute in 

W 

p=-o- (19) 



PISTON TRAVEL. 35 

The distance s traversed by cross-headjpin £ from the beginning 
X of its stroke (see Fig. 14) is equal to 

s = XE= XC — EC ={R^ L) — {Lcosa-\-R cos w) 

s = R{i — cos o>) -f Z(i — cos a). (20) 

This equation holds for both strokes, provided we estimate co from 
0° to 360° and s from the same dead point (left hand dead point 
X, Fig. 14). But if we estimate co and s from the opposite 
center or dead point for the return stroke, and this is commonly- 
done, we must use 

.y = R{i — cos CO) — Z(i — cos a). 

The cosine of the angle made by connecting rod with axis of 
cylinder can be obtained as follows 

J)^ z= L sin o. ^ R sin (o 

hence sin a = — sin co (21) 

I R^ R^ 

and cos a = ^i — — sin^co = i — ^— sin"" co very nearly [(22) 

Introducing this value of cosine a in equation (20) we obtain 

s = r\{ij— cos co) + ^- sin" ^^ J (23) 

For the return stroke, and estimating s and co from the right 
hand dead point this equation becomes 

5 z= 7? (i — cos co) — j4j sin= CO 

From this general formula for the travel, s, of the piston we 
can obtain the velocity, zv, of the piston or of the point E when 
the crank has a uniform velocity. 

According to Elementary Mechanies we have 

for linear velocity, w ^=z —- 

at 

and for acceleration, 6 = —- 

dt 



3^ PISTON VELOCITY. 

Now from (23) 

ds = i^(sin a> + i^ _ sin 2co)do) 

therefore 

2£/ =r i^^sin ^o + i^- sin 2o> j-^ (24) 

is velocity of piston. 

If the crank pin has a uniform circumferential velocity v we 
will have 

Rdco = vdt or — - = — . (25) 

Substituting this value of — - in (24) we obtain 

w = z/fsin 10 + }4- sin 2ft> j ; (26) 

For return stroke and estimation o{ s, co and w from the right 
liand dead point, we have instead 

w ^ vi sin CO — ^— " sin 2^0 j.* 

Differentiating (26) we have 

dw = vlcos wdco -\- y^— cos 2co 2do)j, 

reducing and dividing both members by d^ we obtain 

dw ( , R \dco 

-—= v[ cos CO 4- — cos 2co)—-' 
dt V ' Z ) dt 

Now since the acceleration of a unit of mass is ^ = — - we 

dt 

can substitute d in preceding equation, getting 

fc^ zr= z/l cos a; -j- — cos 2co )-—, (27) 



* An exact expression is 



I R sin 2CM — I 

^1 sin w ± ij — j I 



For exact ratios of w x.o v see Table VIII, p. 60. 



PISTON ACCELERATION. 37 



but we found from (25) that -— = - therefore, 
^ ^' (it R 



- cos ct> -f - cos 2co .* (28) 



For return stroke, and estimation of s, w, co and d from the 
right hand dead point, this becomes 



(' 



R 



d = —( cos w cos 2C0 J . 

At this point we should note that the factor — is identical with the 

R 

acceleration of a unit of mass rotating in the crank-pin circle, and 

when the connecting rod L is infinite the value of 6 for w =0 and 

CO = 180° is ^ = zh -, that is with infinite connecting rod the ac- 
celeration of reciprocating parts, when the crank is on the dead 
centers and iV, is identical with that possessed by a unit of 
mass rotating in the crank pin circle. But when the connecting 

rod is of finite length, d is no longer equal to -, either on dead 

R 

center or on N, but we have instead when co =: o 

.=!(. + f) (.,) 

and when co = 180° 

» - K' - f) w 

* An exact expression for d is : 

V* I sin2 oj L^ C0S2 w — 1 

61 = — I cos OJ =F — . =r ± -5^ — r-^ 

Table VII on p. 39 was not computed from this formula, but from brack- 
eted part of equation 28. The table, to be perfectly exact, should have 
been computed from the bracketed part of the equation just given. The 
difference is so slight, however, that it can be neglected in the problems 
arising in practice. The upper signs correspond to forward stroke. 



38 ACCELERATING FORCE. 



the numerical mean of these two values is, however, equal to -p.- 
we call this mean 6,,,, or 



6. 






1 / rx \ 

X -^ f COS Lo -\- — COS 2C0 j (34) 



(31) 

2 R 

Returning now to the expression deduced above for d and 
substituting in 

W 

F=Md = ^d (32) 

we get /^ = —( cos co -\- - cos 2w j . (33) 

Dividing both members of the equation by A, area of piston 
in square inches, we get 

^_ f^ X '"' X ' ^ ^ 

If the weight W of the reciprocating parts were rotating in 
the crank pin circle, the centrifugal force would be 

P^-'^'R-^-R^ (35) 

. F 

substituting this in the above expression for — we get 

F r . R \F^ F^ 

A 

For return stroke, and estimation of s, co, w, and d from the 
right hand dead point, this becomes 

= I cos 0) ;.- cos 20J )-—=:= C-— . 

^ V L JA A 

It is evident that the value of c will be the sarne numerically for 

angle w of the forward stroke and angle 180° — co for the return 

stroke, but its algebraic sign will be different in these two cases. 

Table VII on following page contains values of c for different 

crank angles co and different values of — • 

K 



F r R \FF 

- = (^cos ^0 -^~ cos 2coy^=: c~J (36) 



TABLE OF ACCELERATION COEFFICIENTS. 



39 



TABLE VII. 



Acceleration of Reciprocating Parts 



Table Containing Values of 



-( 



'A 



R 



cos CO ± J COS 20J. 



The \ "PP^'' \ signs in formula relate respectively to \ ^^^^^'^^^ \ stroke. 
f lower ) ( return ^ 

The algebraic signs in the table relate to forward stroke only ; for return 

stroke, signs opposite to those there given must be used. 



Crank Angles, i 


1 
Connecting-rod h- Crank = 


Crank 


Angles. 


Connecting-rod -=- Crank = 

1 1 


For- 
ward. 


Return. 


4 1 4.5 


50 


5.5 


6.0 


For- 
ward. 


Return. 


I 4.0 


4.5 


5.0 5.5 


6.0 


0°" 


180° 


:i.250 1.222 


1.200 


1.182 


1.167 


90° 


90° 


-.250 


-.222 


-.200 -.182 


-.167 


3° 


177° 


'1.248 


1.220 


1.198 


1.180 


1.165 


! 93° 


87° 


:-.301 


-.272 


-.250 -.233 


-.218 


6° 


174° 


1.240 


1.212 


1.191 


1.173 


1.158 


; 96° 


84° 


-.350 


-.322 


-.301 


-.283 


-.2681 


go 


171° 


1.226 


1.199 


1.178 


1.161 


1.141 


99° 


81° 


-.394 


-.367 


-.346 


-.329 


-315! 


120 


168° 


'1.206 


1.181 


1.161 


1.144 


1.130 


102° 


78° 


,-.436 


-.411 


-.391 


-374 


-3601 


15° 


165° 


11.183 


1.158 


1.139 


1.124 


1.110 


105° 


75° 


-.476 


-.449 


-.430 


-.415 


-.4011 


18° 


162° 


1.153 


1.131 


1.113 


1.098 


1.086 


1 108° 


72° 


-.511 


-.489 


-.471 


-.456 


-.444 


21° 


159° 


,1.120 


1.099 


1.083 


1.069 


1.058 


! 111° 


69° 


-.544 


-.523 


-.516 


-.491 


-482 


24° 


156° 


.1.081 


1.063 


1.048 


1.036 


1.026 


: 114° 


66° 


-.574 


- 556 


-.541 


-.529 


-.519 


27° 


153° 


1.038 


1.022 


1.009 


.998 


.989 


, 117° 


63° 


-.601 


-.585 


-.572 


- 561 


- 552 


30° 


150° ; 


1 .991 


.977 


.966 


.957 


.949 


120° 


60° 


-.625 


-.611 


-.600 


-.591 


-.583 


33° 


147° 


1 .941 


.929 


.920 


.913 


.907 


123° 


57° 


1-.647 


-.635 


-.626 


-.619 


-.616 


36° 


144° : 


1 .886 


.878 


.871 


.865 


.861 


126° 


54° j 


;-.665 


-.657 


-.650 


-.644 


-.640 


39° 


141° 


1 .829 


.823 


.819 


.815 


.812 


1 129° 


51° 


-.681 


-.675 


-.671 


-.667 


-.664 


42° 


138° 


: .769 


.766 


.764 


762 


.760 


i 132° 


48° 


-.695 


-.692 


-.689 


-.688 


-686 


45° 


135° 


' .707 


.707 


.707 


.707 


.707 


1 135° 


45° 1 


-.707 


-.707 


-.707 


-.707 


-.707 


48° 


132° 


.643 


.646 


.648 


.650 


.652 


1 138° 


42° ; 


-.717 


-.720 


-.722 


-.724 


-726 


51° 


129° , 


1 .577 


.583 


.587 


.591 


.594 


i 141° 


39° 


-.725 


-.731 


-.735 


-.739 


-.742 


54° 


126° ! 


.511 


.519 


.526 


.532 


.536 


i 144° 


36° 


-.732 


-.740 


-.7A7 


-.753 


-.758 


57° 


123° . 


.443 


.455 


.464 


.469 


.477 


' 147° 


33° 


-.737 


-.749 


-.758 


-.765 


-.771 


60° 


120° 


.375 


.389 


.400 


.409 


.417 


I 150° 


30° 1 


-.741 


-.755 


-.766 


-.775 


-.783 


63° 


117° 


.307 


.323 


.336 


.347 


.355 


153° 


27° 


-.744 


-.761 


-773 


-.784 


-.793 


66° 


114° 


.240 


.259 


.273 


.285 


.296 


156° 


24° 


-.747 


-.765 


-.780 


-.792 


-.803 


69° 


111° 


.172 


.193 


.209 


.223 


.234 


159° 


21° 


-.748 


-.769 


-.785 


-.798 


-.810 


72° 


108° 


.107 


.129 


.147 


.162 


.174 


162° 


18° i 


-.749 


-.771 


-.789 


-.804 


-.816 


75° 


105° 


.032 


.067 


.086 


.102 


.115 


165° 


15° 


-.750 


-.774 


-.793 


-.809 


-.822 


78° 


102° 


-.020 


.005 


.025 


.032 


.056 


168° 


12° 


-.750 


-.775 


-.795 


-810 


-.826 


81° 


99° 


-.082 -.055 


-.034 


-.017 


-.003 


171° 


go 


-.750 


-.777 


-.797 


-815 


-.830 


84° 


96° 


-.140-. 112 


-.090 


-.073 


-.058 


174° 


6° ' 


-.750 


-.778 


-.799 


-.817 


-.832 


87° 


93° , 


-.197 -.169 


-.147 


-.139 


-.114 


, 177° 


3° i 


-.750 


-.778 


-.800 


-.818 


-.833 


90° 


90° 1 


-.250 -.222 


-.200 


-.182 


-.167 


1 180° 


0° i 


-.750 


-.778 


-.800 -.818 


-.833 



/7 

From the above expression for - we can deduce an approxi- 

A 

mate, graphical, method suf^ciently accurate for most purposes. 

If we make (o zz= 90° and 270° we get, respectively, 

^ ''^" ■ + T^°; (37) 



Z A ^"' 



L A 



40 APPROXIMATE PISTON ACCELERATIONS. 

for CO z= 0° we have 






A 



and for w = i8o° we have 
F 



A \ l) A 



(38) 
(39) 



F^ 



Suppose that we now draw a circle with radius CG = -- and 

A 

lay off CB,CB' and C'B" each equal to ^ and then through 




Fig. 15. 

the three points B',B,B" thus found pass the circle B'BB" the 

middle of the chord B' B will always fall on CO and the 

center of the circle on GBFJ^ Then the acceleration for 

F 
any crank angle FCM will be represented by MN = — and 

j± 

F 
the retardation for the crank angle FCO by ON - • For any 

/{. 

* The center F of the arc B^BB'' falls on the circumference C MC only 

, R 
when -J — \. 



EXACT CONSTRUCTION OF PISTON ACCELERATION. 



41 



Other crank angle the horizontal line included between the two 
circles will measure the corresponding accelerating or retarding 
force. 

Exact Graphical Method of Finding Accelerating of 

Piston and of any Point of the Rod. 
The point Pin Fig. 10 is the instantaneous center for the rela- 
tive motion of connecting rod to engine bed. The velocity of 
crank pin B is to that of slide A as the instantaneous radius PB 
G 9^-.. 




Fig. 16. 

is to radius PA. As each of the radii is also at right angles to 
the direction of the velocity they may respectively be taken as 
the representatives of the velocities of points B and A. As 
velocities may be combined and resolved like forces, we may re- 
gard APB as a triangle of velocities and either side as the resul- 
tant of the other two. Thus PA may be considered as the resul. 



42 EXACT CONSTRUCTION OF PISTON ACCELERATION. 

tant of PB and BA. That is, the velocity of point A may be 
considered as the resultant of the velocity of point B about P 
and of the velocity of point A about ^ as a center (the angular 
velocity about B being then equal to that of B around /*) * 

If we suppose the angular velocity of pin B about center C to 
be uniform and equal to unity, the crank radius CB will represent 
the velocity of its pin and, to the same scale, the distance CD^ 
the velocity of the slide A at the same instant. Then the sides 
of triangle BCD will replace the sides of BPA as the representa- 
tives of the velocities. The length DB thus represents on our 
scale the aforesaid velocity of point A of the rod when turning 
about its point B, 

We are now ready to determine the acceleration of point A. 
When the direction of an acceleration and the intensity of one 
of its two components and the direction of both are known, we can 
easily find the intensity of the acceleration itself by means of the 
triangle of accelerations, (which triangle is analogous to that of 
velocities or forces. In this case we know the direction of the accel- 
eration of the point A to be along its path CAL. We may suppose 
it to be resolved into two components, one along the rod AB and 
the other at right angles thereto. We will determine the one 
along the rod AB. It was shown above that the motion of A 
was compounded of its own motion about B and of ^'s motion. 
Now when the crank has a uniform angular velocity equal to 
unity its acceleration is represented in direction and intensity by 
BC. If we resolve BC into components along and at right 
angles to BA^ the one along the rod will be BC . The part of 
the acceleration of A which is due to its rotation about ^ as a 
center is also made up of two components, one along AB and the 
other at right angles to AB. We need to know only the former, 

* This can also be regarded as an example of the resolution of a rotation 
about one axis P into an equivalent rotation about another parallel axis B 
plus a circular translation. See Weisbach- Herrmann's Machinery of Trans- 
mission. Vol. Ill, Sect. II, §4 of the Introduction. 



EXACT CONSTRUCTION OF PISTON ACCELERATION. 43 

and this is ^ ^J-^. But it was shown above that this velocity 

AB 

= BD, hence the component along AB is equal to -— — . This 

Al:> 

quotient can be found geometrically by describing a circle on 

the rod AB as a diameter and finding its intersections with an 

arc struck from ^ as a center and BD as a radius. The common 

chord 7/ joining these intersections will cut from BA the distance 

BD'^ 
BF^ = — — and will also cut from the horizontal through C, a 
AB 

distance 67/ exactly equal to the desired total acceleration of 

slide A. This can now be easily shown, for the component BF' 

just found has a direction from F^ to B and the component BC^ of 

B along the rod has evidently the direction B to C Therefore i5 (7' 

— F'B = PC = that one of the two components of the total 

accelaration of A which acts along rod AB. Through C draw 

CF equal and parallel to OF^ and to it at i^ erect a perpendicular 

FH; this cuts from ACB^, the direction of the total acceleration 

of A, a distance HC equal to the direction and intensity of the 

total acceleration of slide ^4.* 

The point F' can also be found by drawing through D, the 

intersection of the rod AB with CY the perpendicular to the 

stroke, the parallel BE and where this meets the crank draw £^ ' 

parallel to CY\ this parallel cuts the rod or its prolongation in 

the desired point F . The proof of this construction is : 

BP:BD = BE:BC= BD.BA, that is, BF = — . 

AB 

* This construction is not only the simplest, but it is also applicable to all 
slider-crank mechanisms, whether the stroke of the slide A does or does not 
pass through the crank center C. In the latter as in the former case the 
intercept must be taken on a line C// through C and parallel to the slide 
stroke. It is also applicable to variable crank-pin velocity when scale is such 
that latter is represented by crank AB. We have then only to draw a par- 
allel to stroke through the end of crank-pin acceleration (which in this case 
does not fall at center C of shaft). On this parallel the intercept included 
between the end of acceleration and chord //, Fig. 14, will be the exact 
acceleration of slide A. 



44 METHOD OF FINDING CENTER OF ACCELERATION. 

But this construction of F' evidently fails when the crank is at 
either dead point. 

The determination of the exact acceleration of ^ will enable us 
to construct its curve of acceleration NOQ, which is drawn on the 
stroke of ^ as a base. This suffices for most of the cases arising in 
practice. Greater accuracy is of course obtainable if we ascertain 
the acceleration of each point of the rod, multiply this by the 
elementary mass at that point and then find the resultant of all 
these elementary forces of inertia. This resultant can be resolved 
into two components, one acting at wrist pin A and the other at 
crank pin B. These can then be combined with the other forces 
acting at A and B and the exact tangential pressure on crank pin 
found. We will postpone this combination of forces for the 
present and confine ourselves to finding an easy method of con- 
structing the acceleration of each point of the rod and its com- 
ponents parallel to, and perpendicular to, the rod. 

We will first show that the end of the acceleration KM of any 
point ^of the center line of the rod hes on the straight line joining 
the ends ^'and L of ^5 6" and AL, the accelerations of the two points 
B and A. To do this we must make use of the properties of 
the center of acceleration 6^.* This center is that point of the 
moving rod which has no acceleration. The acceleration of any 
point K is directly proportional to the distance GK of this point 

* For method of finding the center of acceleration G when the directions 
of the acceleration of two points are given and also their ratio, see Weis- 
bach-Herrmann's Machinery of Transmission, Vol, III., Section I, § 21 • 
But for this, most common case, in which the crank has uniform rotation, 
we would suggest connecting crank-pin center B with point H (the inter- 
section of line 1/ with horizontal CB^ Fig. 14) and on this connecting line 
laying off from B a distance equal to the length of connecting rod ; then 
through the end of this distance draw a horizontal till it cuts the prolonga- 
tion of the crank. The triangle thus obtained will be similar to CBH and 
when this triangle is swung around B till its base coincides with BA its 
vertex will fall at G and give exactly the position of the center of acceler- 
ation. The modification necessary for cases of variable rotation of crank 
are evident; the triangles must be similar. 



ENDS OF ACCELERATION FORM AN IMAGE. 45 

from the center of acceleration G, and for the instant in question 
the acceleration of each point makes the same angle with its 
own instantaneous radius of acceleration. For example, let us 
suppose the center of acceleration G known, and let BCy AL and 
KM represent respectively the accelerations of points B^ A and K 
in direction and intensity. 

Then will EC: AL: KM = GB : GA : GK 
and angles CBG = LAG = MKG. 

From these properties, it follows that the triangles GBC, GKM 
and GAL are all similar and hence their corresponding angles ' 
BGC, KGM 2ind AGL are equal to each other. If we turn the 
whole system of points AKB about the center G through the 
angle BGC= KGM ^ AGL, the points B, A^and A will fall 
on B\ K' and A' . Now if we can show that MC is parallel to 
K'B' and ML to K' A' , we shall have proved that MC and ML 
are one and the same straight hne because KB' and K A' con- 
stitute one straight line. From the similarity of the triangles 
GBC, GKM?ind GAL, we have 

GC: GM ^ GB : GK = GB' : GK 
GM: GL =. GK: GA = GK : GA' 

and therefore MC and ML are respectively parallel to KB' and 
K'A' and we have proved that the ends of all the accelerations of 
the points of one straight line lie on one and the same straight 
line. 

This is true of any line of the system and hence of any com- 
bination of lines belonging to or constituting the system. More- 
over it is evident that any hne in the plane of rotation will be 
reduced in the ratio of GC to GB so that 

CL ^ GC 
BA~ GB 

The ends of the accelerations of all points in the plane figure 
constituting the rod will form a reduced image of the shape of 
the rod which will be exactly similar (in the plane of motion) to 
the original rod. This similarity extends to every detail. 
4 



46 ACCELERATION OF ANY POINT ON ROD AND ITS COMPONENTS. 

A corollary from this proposition is that the ends of the 
accelerations of the various points divide the distance CL in the 
same ratio as do their corresponding points the rod or distance 
BA = B'A'. For instance, we have 

CM: ML = B^K'-.K'A' = BK.KA. 

This corollary enables us to determine the acceleration of any 
point of the rod AB (without the help of the center of accelera- 
tion G) when we know the direction and intensity of the accele- 
rations of any two points on that line. For example, if K is half 
way between A and B, the end M of its acceleration KM will be 
half way between the ends C and L of the known accelerations 
BC^xid^AL. 

If the total accelerations of each of the points of the rod be 
resolved into two components, one parallel and the other perpen- 
dicular to the rod, and these components be laid off as ordinates, 
(taking the rod as an axis of abscissas), one set of components on 
one side of the rod and the other set on the opposite side of the 
rod, then will the components of each set terminate on the same 
straight line. That is, the components of the set which are at 
right angles to AB will terminate on the straight line A" B" , and 
the components of the set which coincide with AB, will (when 
revolved through 90°) terminate on the straight line RST. 

This can be proved as follows : Drop from C, J/ and L the re- 
spective perpendiculars CO , MM' and LL' . Because they are a 
series of parallels they will divide the distance C L' in the same 
ratio as 6Z, and since CL has been shown to be divided by the 
acceleration in the same ratio as BA, it follows that C L' is divided 
in the same ratio as BA. But the ordinates C C, M'M, L'L, etc. 
erected on CL', by construction, terminate in a straight line, 
namely CML. As AF' is equal to L' C we may suppose these 
self-same ordinates to be laid off from the former and then they 
will all terminate on the straight line A"F. Now if we suppose 
the base AF' of this set of ordinates to be stretched till it is equal 
to AB in length, the ordinates remaining equidistant and chang- 



ACCELERATION OF ANY POINT ON ROD AND ITS COMPONENTS. 47 

ing only their position and not their magnitude, then will they 
all terminate on the straight line A"B". For before stretching, the 

ratio -f- =1 a, o{ the difference of any two adjacent ordinates to 
ax 

the difference of their abscissas, was equal to a constant a, which 

is the characteristic of a straight line. After stretching, the ratio 

of these differences is — ^ == ^t:' = a constant also, because -— - 
ax'' ax 

AB 
= -jj^ = a constant for the points of this rod and this posi- 
tion of the mechanism. 

If wc revolve the accelerations BC, KM, AL, etc., through 90*^ 
so that they occupy positions BC^\ KM'" , AL'", etc., these new 
positions will also make a constant angle with their instantaneous 
radii of acceleration, and we may show as before that the points 
L"'M"'C"' lie on a straight line. If from their ends C" , M'" and 
L'" wc drop on AB the perpendiculars C" C" , M"'M", L"'L", 
etc., and erect these as ordinates at the corresponding points of 
the base AB we can show as before that these all terminate on one 
straight line RST. But as these perpendiculars C" C" , etc., arc 
respectively equal to the second set of components BC , etc., the 
second part of our proposition has been established. The forces 
of inertia due to this second set of components do not however 
act transversely on the rod, but longitudinally. Their sum or 
resultant combined with the resultant of all the bending forces 
that are due to inertia will give a total resultant, due to inertia of 
rod, that can be resolved into two forces acting at pins A and B.^ 

Exactly how this resultant is found will be discussed later on ; 
at present the shaded area and area ABTR furnish the compo- 
nents for constructing the acceleration of any point belonging to 
the center line of rod. For this purpose however area ABTR 
would suffice, for if its component be laid off on the rod, say 
KM' = KS, a perpendicular erected at M' will cut the line CA 
in yl/ giving at once AW as the acceleration of the point K. 

* This must not be interpreted as giving the total inertia resistance of rod 

for distributed mass. 



48 



ACCELERATION DIAGRAMS. 



Ill A. 

Acceleration Diagrams. 

Obtain by means of the preceding diagrams or tables, or both, 
the accelerating and retarding forces corresponding : to an as- 

F 
sumed value of --, to a given ratio of connecting rod to crank, 

and to certain crank angles. Then find the corresponding piston 
positions as in the preceding plate. At these positions erect 
ordinates equal to the accelerating or retarding forces the former 
being laid off below, and the latter above, the line of piston posi- 
tions. It is evident from the figure that the acceleration curve 
for one stroke can be obtained from that of the other by revolv- 
ing the latter 180° about the central line. 



Cpng^Rqd .. g 
"Crank" "" 1 



Crank 
Angles. 




Fig. 17. 

Construct on the admission line of each of the six diagrams 
of effective steam pressure on piston, two acceleration diagrams 
(in the manner shown in Fig. \f) which correspond respectively to 

A 



— 16 1. 1 Fo 

= say, 40 lbs. and -- 

2 Ji 



2/2 = say, 50 lbs.; p^ is 



the terminal pressure at the end of the expansion curve. Assume 

F 
also ~ = 30. 
A 



PRESSURE ON CROSS-HEAD PIN. 49 

III B. 

Determination of Pressure on Cross-Head Pin, when 

Weight of Rod and Friction are neglected, but 

Inertia taken into Account by assuming whole 

Mass of Rod and Reciprocating Parts 

concentrated on Piston. 

To obtain this diagram we must combine the diagram of effec- 
tive steam pressures on piston with the diagram of accelerating 
and retarding pressures. Moreover the combination must be so 
effected that the accelerating pressures will be subtracted from 
the effective steam pressures, and the retarding forces added. 
This can be done in two ways : by subtracting and adding the 
ordinates of the acceleration diagram, Fig. 17, to the upper lines 
^^^'and EFG of Figs. 8 and 9, representing the pressures on 
driving side of piston, and getting as a result the horizontal pres- 
sure diagrams VTSFH and G'POLA'DC, Figs. 18 and 19, or 
by subtracting and adding the ordinates of Fig. 17 to the lower 
lines HrEdi.nd A' D' C , Figs. 8 and 9, representing the counter 
pressures, and getting; as a result the horizontal pressure diagrams 
A'B'CMNzxid E'FG'HKE\ Figs. 20 and 21. A comparison 
of the two sets of diagrams will show that the corresponding 
ordinates included by the shaded portions are equal. 

The shaded area in each case is bounded on the one side by 
the resultant line (due to the combination of expansion line or 
line of counter-pressure with acceleration curve) and on the 
other side by that line of Fig. 8 or 9 which did not help to form 
the resultant line. 

It should also be noticed that where the resultant line crosses the 
unused line (given by Fig. 8 or 9) the horizontal pressures on cross- 
head pin become equal to zero, and after passing this point, see 
Fig. 21, the direction of the horizontal pressures on cross head pin 
is reversed before the piston reaches the end of its stroke. In 
Fig. 20 the steam and retarding pressures are so related as to com- 



50 



PRESSURE ON CROSS-HEAD PIN. 




Fig. ig. 

pletely unload the crank-pin at the end of the stroke. This is 
not so favorable as it seems, for at the beginning of the return 
stroke there is suddenly applied a great pressure in the oppo- 
site direction. A little more compression /'5, Fig. i8, would 
have changed this. It is probable that the arrangement shown in 



PRESSURE ON CROSS-HEAD PIN. 



51 




Fig. 20 



Fig. 21, is less likely to cause pounding than that in Fig. 18, and 
this in spite of the fact that in the case of Fig. 21 the reversal of 
pressure near end of stroke occurs while piston is in motion. A 




Fig. 21. 

less lead in the case of Fig. 19 would be advantageous in cutting 
down the large initial pressure. 

Another construction used by Grashof, is well worth noting. 
It is to lay off these resultant pressures at cross-head pin on the 
rectified crank-pin circle. The straight base will then represent 
time, and the rate at which the pressure changes can then be 



52 



EFFECT OF VALVE SETTING ON RESULTANT PRESSURE. 



easily seen. This will be particularly useful in judging of the 
rapidity with which the reversals of pressure occur. 

We shall hereafter only make use of horizontal pressure or 
wrist-pin diagrams similar to those shown in Figs. i8 and 19. 

Ill c. 
Effect of Valve Setting on the Horizontal Pressure 

ON Cross-Head Pin. 
We give two indicator diagrams, Figs. 22 and 23, from opposite 
ends of the same cylinder to show the influence respectively of a late 

A B 




Fig. 22. 

admission and too late an exhaust. By combining the diagram 
ABCUV, Fig. 2 2,with diagrams of counter-pressure HIJK, Fig.23, 
we get the effective steam pressure diagram ABCKJIH, Fig. 24. 

GT F 




Hehirn jStfO^e, 



Fig. 23, 



EFFECT OF VALVE SETTING ON RESULTANT PRESSURE. 53 




In like niamicr by combining the counter-pressure diagram 
EDUV, Fig. 22, with HGFKYZ, Fig. 23, we get the effective 
pressure diagram KDEHG, Fig. 25. By combining these real 



54 EFFECT OF VALVE SETTING ON RESULTANT PRESSURE. 

pressure diagrams with the acceleration diagrams we get the 
horizontal pressures on cross-head pin, shown in Figs. 24 and 25 
by shaded areas. Fig. 24 shows that, owing to the late exhaust 
HI, Fig. 23, there is a drag instead of a push on the crank dur- 
ing the first portion of the stroke, and a reversal of pressure on 
the crank while the piston is in rapid motion at L. This reversal 
will cause a shock or knock if there is any play in the joints of 
the mechanism. In like manner Fig. 25 shows a drag instead of 
a push on crank at beginning of stroke and a reversal of pressure 
at D, but here the piston motion is considerably slower than at 
Z, so there will probably be no shock. The final pressure, 
KS Fig. 25, on piston at beginning of return stroke is about 
equal to that, KN Fig. 24, at end of forward stroke, and is 
a desirable result. It would be better however to accomplish 
the same result by more compression and earlier admission. 
Fig. 25 also shows the very unequal distribution of the horizontal 
pressures which attends a large cut-off when unaccompanied by 
suitable compression. By having the proper amount of com- 
pression at Sy Fig. 18, not only may the reciprocating parts be 
brought to rest without shock, but all load can be taken off the 
crank pin while it is passing the dead center. This is not always 
done on account of the decided advantages gained by compres- 
sing nearly up to the boiler pressure, some of which advantages 
are the filling and heating of the hurtful space by compressed 
steam instead of the live, fresh, steam and the avoidance of such 
admission lines as K,F Fig. 23. Moreover, the reversal of 
pressure which accompanies considerable compression is not 
necessarily an evil, provided it takes place when the piston is 
moving slowly, i. e., near the end of stroke where reversal natur- 
ally takes place. The ideal case is to have the piston unloaded 
while moving slowly (near end of stroke) and then let the 
driving pressure on opposite side increase gradually till after the 
reverse stroke has begun. Considerable compression and even 
a slight negative lead may be used to effect this desirable result, 
desirable, because of its freedom from pounding. 



EFFECT OF SHORT CUT-OFF ON RESULTANT PRESSURE. 



55 




Fig. 26. 



In Fig. 26 the ordinate AH at the beginning of the accelera- 

F 
tion curve -^ is greater than the corresponding effective steam 

pressure AG\ the result is that instead of there being a 
push upon the crank in the direction of its motion at the 
beginning of the stroke, there is a drag upon the crank, this 
condition continuing until the piston reaches the position /, 
where a reversal of pressure takes place causing shock and 
vibrations injurious to the durability of the machine, because 
at that end of the stroke there is considerable piston speed at 
point /. 

In Fig. 27 the condition of affairs is still worse, for two rever- 
sals of pressure take place at /' and I" , causing two shocks in 
rapid succession. In this last case, although the effective pres- 
sure AG is greater than the initial pressure of acceleration A' H' , 



the ordinates 



\ A A^ 



F^, , R 

cos iO -Y — cos 2 CO 



)^ofth 



e acceleration 



56 



EFFECT OF SHORT CUT-OFF ON RESULTANT PRESSURE. 




Fig. 27. 

Fo A' B' 

curve -J are still too large for the short cut-off--- - and the small 

initial pressure A'G' . It is evident from the figures that if the 
effective pressure diagrams ABCDEFGA ^xvA A' B' C D' E F G' A' 
of Figs. 26 and 27 are to remain unchanged, that is, if the 
mean effective pressure pin is to remain unchanged, the only- 
remedy for the evils exhibited in Figs. 26 and 27 is to diminish 

F 
the value of — ?. 
A 

When the engine varies between wide limits of cut-off, as in 

Fo 
the present case, the value of -j is chosen so as to favor as much 

as possible the particular horse-power at which the engine ordi- 
narily runs ; in such a case it is desirable to know what is the 
minimum cut-off at which reversal of pressure will be avoided. 
This may be ascertained by the cut and try method, but it may 



REVERSALS OF PIN PRESSURES DUE TO SHORT CUT-OFF. 



57 



also be calculated with sufficient accuracy if we make the follow- 
ing suppositions, namely, that the curve of expansion is an equi- 

p 
lateral hyperbola, that the best value for -^ is 

K p, — i6 

and that the crank angle is about 60° when the horizontal pres- 
sure curve EFG (Fig. 28), is tangent to the back pressure line 




Fig. 28. 



DF. As the reversals in question are more apt to occur during 
the return stroke we must use a formula 



p — p^ = -^ (cos CO— ~ cos 2w\ =. -^ p^—p 



(40) 



corresponding to that stroke and introduce the assumptions just 
made. 



* Radinger holds that — — 2/2 = twice the terminal pressure at end of 



expansion will give the steadiest running of engine. 



58 CONVERSION OF HORIZONTAL INTO TANGENTIAL PRESSURES. 

By deriving a value for v from the piston travel s for return 
stroke and allowing for clearance, we find that under the average 
conditions of running of high-speed engines, the minimum, real, 
cut-off should be greater than o.io to avoid reversals of pressure 
when piston is running swiftly. The approximate formula for 
this cut-off is : 

W>sk5-(f-)+&=(T«-a(f-) + |,'-' 

To find the number of horse-powers corresponding to this 

minimum, real, cut-ofifwe look in Table III for value of^^— ^ and 
then transforming, Eq. 14, we get 

-[■-'-■<'^/-)]- 



All the terms of the second member being known we can calcu- 
late readily the minimum horse-power at which the engine can 
be safely run at the given speed and pressure. 

Find minimum cut-off and horse power for present engine. 
Construct on each of the six diagrams of effective steam pressures 
on piston two diagrams of resultant pressures on cross-head pin 

F p 16 F F 

corresponding to -^ = — and -° = 2/2, i.e. -^ = 50 and 

40 lbs. 



Ill D. 

Conversion of Horizontal Pressures of Crank Pin into 
Tangential Pressures. 

The horizontal component of the pressure of the rod against 
the crank-pin is evidently equal to the horizontal pressure of 
the cross-head pin. In this problem we are to find the rotative 
effect of a horizontal pressure on crank-pin. 



CONVERSION OF HORIZONTAL INTO TANGENTIAL PRESSURES. 59 

Let P* represent horizontal pressure on cross-head pin. 
T represent force along connecting rod. 
/ represent tangential pressure on crank pin. 

then will T = (42) 

cos a 

and t=T sin {<« + <^) = /" ^'"(^"+ ^ (43) 

COS a 



t sin [to -[- a) 

P^ COS a 



(44) 



If we substitute in this equation sin a = y sin o) and cos a == 



4 



-— sin^ CO (see Eqs. 21 and 22) we shall have the means 
-^2 



of calculating the ratios — from the crank angle co. Table VIII 

on page 60 contains these ratios. 

If we wish to determine the tangential pressures from given 
horizontal pressures by graphical means we can proceed as fol- 
lows, Fig. 29. 

In triangle AKC, Fig. 29, we have 

KC _KC _ sin {o) + a) 
AC R cos a 

or 

t_ _KC _w 



(45) 
(46) 



hence if on any scale, say 20 lbs. to the inch, we lay off Z^^= /*', 
join K with D and prolonging KD till it intersects at F the per- 
pendicular to DE erected at E, we will have in the two similar 
triangles, DKC and DEE, the proportion FE\ P' : : KC: R, i. e. 
t = FE. 



6o 



TABLE FOR CONVERSION OF PRESSURES. 



TABLE VIII. 

Rotative Effect of a Unit of Horizontal Pressure 

ON Crank. 

To find actual tangential pressure (or rotative effect) on crank, multiply 
tabular quantity by resultant horizontal pressure on crank or cross-head pin. 
Forward stroke is towards, and return stroke away from, crank shaft. 
To find wrist-pin velocity multiply tabular values by crank-pin velocity. 



Crank Angles. 


Connecting Rod -^- Crank = 


Forw'd. 


Return. 


40 
.1089 


4-5 


5.0 


5-5 


6.0 


CX3 


5 


175 


.1064 


.1045 


.1030 


.1016 


.0832 


lO 


170 


1 .2164 


.2117 


.2079 


.2047 


.2022 


.1737 


15 


165 


•3215 


•3145 


.3089 


•3054 


.3005 


.2588 


20 


160 


.4227 


.4136 


.4065 


.4019 


•3957 


.3420 


25 


i55 


.5189 


.5081 


•4995 


•4925 


.4866 


.4226 


30 


150 


.6091 


.5968 


.5870 


•5791 


.5724 


. .5000 


35 


145 


.6923 


.6788 


.6682 


.6596 


.6523 


.5736 


40 


140 


.7675 


.7533 


.7421 


.7329 


.7253 


.6428 


45 


135 


.8341 


.8195 


.8081 


.7988 


.7910 


.7071 


50 


130 


.8914 


.8771 


.8657 


.8564 


.8488 


.7660 


55 


125 


.9392 


•9253 


.9144 


•9055 


.8982 


.8192 


60 


120 


.9769 


.9641 


.9540 


.9458 


.9390 


.8660 


65 


115 


1.0046 


•9932 


.9842 


.9769 


.9709 


.9063 


70 


no 


1.0224 


1.0127 


1.0052 


.9990 


■9939 


.9397 


75 


105 


1 .0304 


1.0228 


1.0169 


1.0121 


1 .0082 


.9659 


80 


100 


1.0289 


1.0237 


1.0199 


1. 01 64 


1.0137 


.9848 


85 


95 


1.0186 


1. 0160 


1. 01 39 


1. 01 27 


1. 0109 


.9962 


90 


90 


1 .0000 


1 .0000 


1 .0000 


1 .0000 


1. 0000 


1. 0000 


95 


85 


•9738 


.9764 


.9785 


.9797 


.9816 


.9962 


100 


80 


.9407 


.9460 


.9500 


.9532 


•9559 


.9848 


105 


75 


.9016 


.9091 


.9150 


.9198 


.9237 


.9659 


no 


70 


.8571 


.8667 


.8743 


.8804 


.8855 


.9397 


115 


65 


.8080 


.8194 


.8285 


.8357 


.8416 


.9063 


120 


60 


•7552 


.7680 


.7781 


.7863 


•7931 


.8660 


125 


55 


.6992 


.7130 


■7239 


.7328 


.7401 


.8192 


130 


50 


.6407 


.6550 


.6664 


.6756 


.6833 


.7660 


135 


45 


.5801 


.5946 


.6061 


.6155 


.6232 


.7071 


140 


40 


.5181 


.5323 


.5435 


.5527 


.5603 


.6428 


145 


35 


.4549 


.4683 


.4790 


.4876 


•4949 


.5736 


150 


30 


.3909 


.4032 


.4130 


.4209 


.4276 


.5000 


'|5 


25 


.3264 


•3371 


.3458 


.3528 


.3586 


.4226 


160 


20 


.2614 


.2704 


.2776 


.2821 


.2884 


.3420 


165 


15 


.1962 


.2032 


.2088 


.2123 


.2171 


.2588 


170 


10 


.1309 


.1356 


.1394 


.1426 


.1451 


.1737 


175 


5 


.0655 


.0679 


.0698 


.0714 


.0727 


.0872 



CONVERSION OF HORIZONTAL INTO TANGENTIAL PRESSURES. 6 1 

The value of t will be on the same scale as P^. If we determine 
in like manner, for the same pressure P' ==z DE, the tangential 
pressures corresponding to the various crank positions of Fig. 29 
and lay them off on the extreme left or right hand ordinates of 




Rciurn. 



Fig. 29, 



the section paper employed, then drawing diagonals to the 
assumed zero of cross-head pin pressures we will get a diagram 
similar to Fig. 30 from which we can determine the tangential 
pressures for any value of P'. In Fig. 29 the ratio of length 
of connecting rod to length of crank was taken at 2j/^. In the 
diagram to be drawn on section paper assume this ratio to be 
= 6, also DE= P' = 200 lbs. 

Lay off, in Fig. 30, the tangential pressures on a scale of 20 
lbs. to the inch, and the horizontal or cross-head pressures on 
a scale of 20 lbs. to the inch. The tangential pressures can also 
be obtained in the ordinary manner by resolving the force acting 
along the connecting rod into two components respectively tan- 
5 



62 



DIAGRAM FOR CONVERSION OF PRESSURES. 



gent and normal to the path of the crank pin. But this would 
involve a good deal more labor than the present method. Figs. 
29 and 30 are quite accurately drawn and the method of obtain- 
ing 30 from 29 can therefore be easily followed. Thus EL, EM 
and EN are the tangential pressures corresponding respectively 
to the angular positions 20°, 40° and 60° of crank for return 




200 




Horizontal pressure on Cross-head pin. 

Resultant of effective steam pressure and inertia of reciprocating parts. 

Fig. 30. 



Stroke 2X^6, \.o 160°, 140° and 120° ior \.\\c forward stroke. The 
quickest method of obtaining a diagram like Fig. 30 would be to 
calculate the tangential pressures from Table VIII, assuming 
a constant horizontal pressure of 200 lbs. on cross-head pin. It 
is sufficiently evident from Fig. 30 how the tangential pressures 
for other and smaller resultant pressures on cross-head pin can 
be obtained. 



CONSTRUCTION OF TANGENTIAL PRESSURES. 63 



III E. 

Construction of Diagram of Tangential Pressures on 
Crank Pin. 

We are now ready to construct the tangential pressure dia- 
grams from the diagrams of resultant pressures on cross-head 

Fo 
pin corresponding to various values of ^. We proceed as in the 

construction of Fig. 13. By means of the diagram, Fig. 30, we 
first transform the horizontal pressures into tangential pressures, 
and then lay off these tangential pressures as prolongations of 
the radii of the crank-pin circle (or of any other convenient 
circle), the latter forming the base of the radial ordinates. A 
curve is next drawn through the extremities of these ordinates 
and then a circle concentric with — and outside of — the crank-pin 
circle, is drawn, the difference between the two circumferences is 
equal to the mean tangential pressure, which as before is equal 
to pt = 0.6^66 p,„, Eq. 17. The deviations of the curve (drawn 
through the extremities of the ordinates representing the tan- 
gential pressures) from the circle of tangential pressures will 
show the irregularity of the driving power. The more closely 
the tangential pressure curve approximates to the mean-tan- 
gential-pressure-circle, the steadier will the engine run. Figs. 
31 and 32 show two different methods of representing the 
tangential pressures. In the former figure the tangential pres- 
sures are laid off from the crank-pin circle itself; in the latter 
from the rectified semi-circle. Fig. 31 has the advantage of 
showing the phases of resistance CMNO and AQS and the curve 
of tangential pressures in a continuous manner, but it is subject 
to the following very slight disadvantage, viz : that the area en- 
closed by AQRMCBA does not represent with great accuracy 
the work done in a semi-revolution, the inaccuracy being due to 
the divergence of the radial ordinates. The objection does not 
apply to Fig. 32, where the area A VECBA is exactly equal to the 



64 



CONSTRUCTION OF TANGENTIAL PRESSURE DIAGRAMS. 



area AF'DCA = work done in one semi-revolution ; moreover 
by adding the shaded portions CiyE' {=CI)B) and APG' as 
shown in Fig. 32, the objection that this mode of representation 
does not represent the phases of resistance in a continuous man- 




^^>! 



Fig. 33. 

ner disappears. If the work has been done correctly it will be 
found that in Fig. 3 1 the area of the shaded portion QRM = 
area CMN + area QAS, nearly, and that in Fig. 32 the phase of 
excess GVE =- AGF + /)£(7 exactly. 

Inspection of either of these figures shows that the tangential 
pressure line (shown by full line) represents steadier running than 



USE OF TANGENTIAL PRESSURE DIAGRAMS. 65 

the tangential pressures corresponding to greater piston speed or 
heavier reciprocating parts (shown by broken line). 

Draw for the minimum, rated and maximum power of the pro- 
posed engine, separately two tangential pressure diagrams, one 
like Fig. 31 and the other like Fig. 32. On each diagram draw 
two tangential pressure curves corresponding to the resultant, or 
horizontal, pressure on cross-head pin which were drawn for the 

p p^ i5 p^ 

assumed values: -^ = — = about40 and-^°=: 2/2= about 

A 2 A 

50 lbs. When these have all been drawn, inspect the result and 

p 
see if a different value than those chosen for — would not give a 

more uniform tangential pressure — i. e. steadier motion. If so, 

take the new value of ~ and draw corresponding acceleration 

A 

curves for both the forward and the return stroke, and superim- 
pose these upon the diagrams of effective steam pressures already 
drawn and combine them as before to form diagrams of resultant 
pressures on cross-head pin. Then convert these last diagrams 
into tangential pressure diagrams and again inspect the latter 
diagrams to see if any improvement can be effected. This pro- 

p^ 
cess is to be continued till that value for ~ has been found which 

A 

gives the most uniform tangential pressure for the normal, rated, 
horse-power of the proposed engine. This most favorable value 

of ~ for the rated power should however be examined with refer- 
ence to the minimum power required, to see whether it then 
causes reversals of pressure. When this is the case some smaller 

p^ 
value of -^ less favorable to the normal or rated power should be 

chosen for the engine. In this connection we may quote the 
following from Rigg's Steam Engine. 



66 



CHOICE OF SPEED AND LOAD. 



Choice of Speed and Load. 



" There are, then, three elements which can be adjusted to each 
other, namely, steam pressure, including the rate of expansion, 
weight of reciprocating parts, and number of revolutions per 
minute, or piston velocity ; any of these elements can be altered 
at the option of the designer of the engine, and the problem to 
be solved is not to transfer the strains from one end of the stroke 
to the other, not to work steadily with the highest expansion, not 
to produce a regular uniform horizontal pressure on the crank. 
But the end and object of all these calculations and changes in 
the old acknowledged rules for the construction of engines is, by 
working with that amount of expansion which practical ex- 
perience proves economical, to obtain as nearly as possible a uni- 
form tangential pressure on the crank. When this is done an 
engine will drive its load steadily and well, and the influence of 
the reciprocating parts has a most direct bearing on this most 
important subject. 

In applying the foregoing reasoning it is necessary to exercise 
a judicious choice in balancing the evils which arise from a too 
low or too high speed, and so to decide a rate at which an engine 
can be run to the best advantage. When the load upon the 
engine is regular it is comparatively easy to do this, but in a great 
majority of cases the load is often varying, and the speed is bound 
to remain constant. Thus, it is necessary to select a load on the 
engine which shall meet most requirements, and then the indica- 
tor diagram either actually taken or assumed will form the prin- 
cipal datum required." 

To these considerations might be added another, namely that 
to avoid pounding or shocks at crank- or wrist-pin, the driving 
pressures should change as gradually as possible from one side to 
the other of piston, the change taking place as near the dead 
point as possible. By using considerable compression or even a 
slightly negative lead this gradual change may be accomplished 
for both ends of the stroke. 



DETERMINATION OF DIAMETER OF CYLINDER. 67 

IV. 

Determination of Diameter of Cylinder D, Length 

OF Stroke 5, Revolutions per Minute N or 

Piston Speed per Minute w"- and Weight 

OF Reciprocating Parts W by Means 

OF the Tables or Formulas. 



A 


W W 
= 0.0000142 -irSN^ = .000085 —Nw"- 
A. A 




Fo 
'A 


, H WN ^ w' W H 
-^■^^p-rAD' = ''-'si> Ap- 


(47) 




TT 

— == 0.0000238 W^I> 

Pm 


(48) 




— = 0.00000396 NSD^ 

pm 


(49) 



^ = 3.63^^ = 21.8^5^ (50) 

pm 

H = number of indicated H. P. 

5 = length of stroke in inches. 

IV . . 

— -= weight ot reciprocating parts per □'' of piston, weight 

A 

of rod included. 

D = diameter of cylinder in inches. 

w^ = average speed of piston in feet per minute. 

N = revolutions per minute. 

P^ = mean effective pressure per Q" of piston. 

K 

~ = average force accelerating reciprocating parts when 

A 

crank is on dead centers. 



68 



MEAN EFFECTIVE PRESSURE ON PISTON. 



TABLE IX.* 



it- 
Value OF — OR Indicated H. P. for each Pound of Mean 

■ Effective Pressure per □" of Piston Area. 

Quantities in table were obtained by neglecting area of piston rod and 
assuming that the same mean effective pressure {pm ) existed on both sides 
of the piston area ; consequently the tabular quantities are about i Yz % and 
Yz % too large for small and large cylinders respectively. To correct for 

piston rod multiply tabular quantities by [i — Y \n/ 1 where B = diameter 
of cylinder and d^ = diameter of piston rod. 



H 

P u 


i 
1 




Average speed of 


piston in feet 


per minute. 




1 700 
.600 


750 
■ .643 


800 

.685 


850 


900 


950 


1000 


1050 


1100 


1150 


1200 


6 


.728 


.771 


.814 


857 


.900 


.942 


.985 


1.028 


6^ 


.704 


.754 


.804 


.855 


.905 


.955 


1.006 


1.056 


1.106 


1.156 


1.207 


7 


.816 


.875 


.933 


.991 


1.050 


1.108 


1.116 


1.225 


1.283 


1.341 


1.399 


7% 


j .937 


1.004 


1.071 


1 138 


1.255 


1.272 


1 339 


1.406 


1.473 


1.540 


1.607 


8 


1 1.066 


1.142 


1 219 


1.295 


1.371 


1.447 


1.523 


1.599 


1.676 


1.752 


1.828 


8^ 


1 1.204 


1.290 


1.376 


1.462 


1.548 


1.634 


1.719 


1.806 


1.892 


1.977 


2 063 


9 


1 1 349 


1.446 


1.542 


1.639 


1.735 


1.831 


1.928 


2.021 


2.121 


2 217 


2.313 


9^ 


1 1.504 


1.611 


1.718 


1.826 


1.933 


2.041 


2.148 


2 255 


2.363 


2.470 


2.578 


10 


j 1.666 


1.785 


1 904 


2.023 


2.142 


2.261 


2 380 


2.499 


2.618 


2.737 


2.856 


10^ 


j 1.874 


2.008 


2.142 


2.276 


2.420 


2.544 


2.678 


2.811 


2.945 


3.079 


3.213 


11 


1 2.016 


2.160 


2.304 


2.448 


2.592 


2.736 


2.880 


3.024 


3.168 


3.312 


3.456 


ny2 


1 2.203 


2.361 


2.518 


2.675 


2.833 


2.990 


3.148 


3.305 


3.462 


3.620 


3.777 


12 


2.399 


2.570 


2.742 


2.913 


3.084 


3.256 


3.427 


3.599 


3 770 


3.941 


4.113 


13 


2.816 


3-017 


3.218 


3.419 


3.620 


3.821 


4.022 


4.223 


4.424 


4.626 


4.827 


14 


3.265 


3.500 


3.732 


3.965 


4.198 


4 432 


4.665 


4.896 


5.231 


5.365 


5.597 


15 


3.749 


4.016 


4.284 


4.552 


4.820 


4 937 


5.355 


5.623 


5.891 


6.158 


6.426 


16 


4.265 


4.570 


4:774 


5.179 


5 484 


5.788 


6.093 


6.397 


6.696 


7.007 


7.311 


17 


4.815 


5.159 


5.503 


5.846 


6 190 


6.534 


6.878 


7.222 


7.566 


7.810 


8.254 


18 


5 397 


5.783 


6.169 


6.555 


6.940 


7.. 326 


7.711 


8 087 


8.482 


8.868 


9.253 


20 


6.664 


7 140 


7 616 


8.092 


8.568 


9.044 


9.520 


9.996 


10.472 


10.948 


11.424 


22 


1 ,8.063 


8.639 


9.215 


9.791 


10.367 


10.943 


11.519 


12 095 


12.577 


13.147 


13.823 ! 


24 


1 9.596 


10 282 


10.967 


11.652 


12.338 


13.023 


13.709 


14.394 


15.080 


15.765 


16.451 


26 


11.162 


12.067 


12.871 


13.675 


14.480 


15.284 


16.089 


16.893 


17.998 


18.502 


19.307 


28 


1 13.061 


13.984 


14.927 


15.860 


16.793 


17.726 


18.659 


19.592 


20.525 


21.458 


22.391 


30 


1 14.994 


15.065 


17.136 


18.207 


19.278 


20.349 


21.420 


22.491 


23.562 


24.633 


25.704 


32 


1 17.060 


18.278 


19.497 


20.716 


21.934 


23.153 


24.371 


25.590 


26.708 


28.027 


29.245 


34 


19.259 


20.635 


22.010 


23.386 


24.762 


26.137 


27.513 


28.888 


30.264 


31.640 


33.035 


36 


21.591 


22.134 


24.676 


26.218 


27.760 


29.303 


30.845 


32.380 


33.929 


35.472 


37.014 


38 


24.057 


25.775 


27.494 


29.212 


30.930 


32.649 


34.367 


36.086 


37.804 


39.532 


41.339 


40 


! 26.656 


28,560 


30.464 


32.368 


34.272 


36.176 


38.080 


39.984 


41.888 


43.792 


45.696 



* Table IX continued on page 69. 



MEAN EFFECTIVE PRESSURE OF PISTON 



69 



IS 

Value of — or Indicated H. P. for each Pound of Mean 

Pni 

Effective Pressure per □'' of Piston Area. 

This part taken frotn Riggs' Steam Engine. 



M 

4 


, Average speed of piston in feet per minute. 




1 

240 

.091 


I 300 
.114 


! .133 


400 


1 450 


500 


1 550 


600 


650 


750 




1 .162 


i .171 


.19 


' .209 


.228 


.247 


.285 




4J^ 


.115 


! .144 


1 .168 


i .192 


.216 


.24 


i .264 


.288 


.312 


.36 




5 


.144 


18 


i .21 


1 .24 


1 .27 


.30 


.33 


.36 


.39 


.45 




^% 


.173 


.216 


i .252 


i .288 


.324 


.36 


.396 


.432 


.468 


.54 




6 


1 .205 


.256 


! .299 


.342 


.385 


.428 


1 .471 


.513 


.555 


.641 




6J4 


.245 


307 


j .391 


.409 


.461 


.512 


.563 


.614 


.698 


.800 




7 


.279 


.348 


! .408 


.466 


.524 


.583 


1 .641 


.699 


.756 


.874 




7J^ 


i .321 


.401 


.468 


.534 


.602 


.669 


.735 


.802 


.869 


1.002 




8 


' .365 


.456 


.532 


.608 


.685 


.761 


1 .837 


.912 


.989 


1.121 




8'.j 


.413 


.516 


.602 


.688 


.774 


.86 


! .946 


1.032 


1.118 


1.29 




9 


.462 


.577 


.674 


.770 


,866 


.963 


1.059 


1.154 


1.251 


1.444 




9J^ 


.515 


.644 


.751 


.859 


.966 


1.074 


1.181 


1.288 


1.395 


1.610 




10 


.571 


.714 


.833 


.952 


1.071 


1.190 


1 1.309 


1.428 


1.547 


1.785 




lOj^ 


.63 


.787 


.919 


1.050 


1.181 


1.313 


1 1.444 


1.575 


1.706 


1.969 




11 


.691 


.864 


1.008 


1.152 


1.296 


1.44 


1.584 


1.728 


1.872 


2.160 




\\% 


.754 


.943 


1.1 


1.257 


1.414 


1.572 


1.729 


1.886 


2.043 


2.357 




12 


.820 


1.025 


1.195 


1.366 


1.540 


1.708 


! 1.880 


2.050 


2.222 


2.564 




13 


.964 


1.206 


1.407 


1.608 


1.809 


2.01 


1.211 


2.412 


2.613 


3.015 




14 


1.119 


1.398 


1.631 


1.864 


2.097 


2.331 


1 2.564 


2.797 


3.029 


3,495 




15 


1.285 


1.606 


1.873 


2.131 


2.409 


2.677 


2.945 


3.212 


3.479 


4.004 




16 j 


1.461 


1.827 


2.131 


2.436 


2.741 


3.045 


3.349 


3.654 


3.958 


4.567 




17 1 


1.643 


2.054 


2.396 


2.739 


3 081 


3.424 


3.766 


4.108 


4.450 


5.135 1 


18 ; 


1.849 


2.312 


2.697 


3.083 


3.468 


3.854 


4.239 


4.624 


5.009 


5.78 




19 


2.061 


2.577 


3.006 


3.436 


3.865 


4.295 


4.724 


5.154 


5.583 


6.442 




20 1 


2.292 


2.855 


3.331 


3.807 


4.285 


4 759 


5.234 


5.731 


6.186 


7.138 




21 ! 


2.518 


3.148 


3.672 


4.197 


4.722 


5.247 


5.771 


6 296 


6.820 


7.869 




22 1 


2.764 


3.455 


4.031 


4.607 


5.183 


5.759 


6.334 


6.911 


7.486 


8.638 




23 ! 


3.021 


3.776 


4.404 


5.035 


5.664 


6.294 


6.923 


7.552 


8.181 


9.44 




24 


3.289 


4.111 


4.797 


5.482 


6.167 


6.853 


7.538 


8.223 


8.908 


10.279 




25 


3.569 


4.461 


5.105 


5.948 


6.692 


7.436 


8.179 


8.923 


9.566 


11.053 




26 


3.861 


4.826 


5.630 


6.435 


7.239 


8.044 


8.848 


9.652 


10.456 


12.065 




27 


4.159 


5.199 


6.066 


6.932 


7.799 


8.666 


9.532 


10.399 


11.265 


12.998 




28 


4,477 


5.596 


6.529 


7.462 


8.395 


9.328 


10.261 


11.193 


12.125 


13.991 




29 


4.805 


6.006 


7.007 


8.008 


9.009 


10.01 


11.011 


12.012 


13.013 


15.015 




1 30 


5.141 


6.426 


7.497 


8.558 


9.639 


10.71 


11.781 


12.852 


13.923 


16.065 




31 


5.486 


6.865 


8.001 


9.144 


10.287 


11.43 


12.573 


13.716 


14.866 


17.145 




1 32 


5.846 


7 308 


8.526 


9.744 


10.962 


12.18 


13.398 


14.616 


15.834 


18.270 




33 


6.216 


7.770 


9.065 


10.360 


11.655 


12.959 


14.245 


15.54 16.835 


19.425 




34 


6.59 


8.238 


9 611 


10.9S4 


12.357 


13.73 


15.103 


16.476 17.849 


20.595 




35 


6.993 


8.742 


10.199 


11.656 


13.113 


14.57 


16 027 


17.484 18.941 


21,855 




36 


7.401 


9.252 


10.794 


12.336 


13.878 


15.42 


16 962 


18.504 , 20.046 


23.130 




37 


7.819 


9.774 


11.403 


13.033 


14.861 


16.29 


17.919 


19.548 ' 21.177 


24.435 




38 


8.246 


10.308 


12.025 


13.744 


15. 462 1 


17.18 


18.896 


20.616 , 22.334 


25.770 




39 


8.648 


10.86 


12.67 


14.48 


16.29 


18.1 


19.91 


21.62 ; 


23.53 


27.15 




40 


9 139 


11.424 


13.328 


15.232 


17.136 


19.04 


20.944 


22.848 ; 


24.752 


28,560 




41 


9.604 


12.006 


14.007 1 


16.008 


18.009 ' 


20.00 


22.011 . 


24.012 : 


25,013 


30.015 ' 


42 i 


10.065 


12.594 


14.693 t 


16.792 , 


18.901 


20.99 


23.089 


25.188 ' 


27.287 


31.485 




43 1 


10.56 


13.20 


15.4 1 


17.6 


19.8 ; 


22.0 


24.2 


26.4 ; 


28.6 


33.0 




44 ' 


11.046 


13.818 


16.121 1 


18.424 ! 


20.727 


23.03 


25.333 


27.636 


29.939 


34.545 




45 


11.563 1 


14.454 


16.863 : 


19.272 i 


21.681 : 


24.09 


26.399 


28.908 


31.317 


36.135 




46 


12.086 j 


15.128 


17.625 


20.144 1 


22.662 ' 


25.18 


27.698 


30.216 


32.754 


37.770 




! '*' 


12.614 


15.768 ; 


18.396 ; 


21.024 1 


23.652 1 


26.28 


28.908 


31.536 1 


34.164 


39.420 i 




1 48 . 


12.846 


16.446 


19.187 1 


21.928 1 


24.669 ! 


27.41 i 


30.151 


32.152 , 


35.633 


41.115 1 




1 49 


12.913 


17.142 1 


19.999 • 


22.856 ■ 


25.713 : 


28.57 


31.427 1 


34.284 ' 


37.141 


42.855 1 




! 50 ! 


14.28 , 


17.85 i 


20.825 ' 


23.8 ; 


26.775 i 


29.75 


32.725 


35.7 , 


38.675 


44.625 




51 


14.832 1 


18.54 i 


21.665 ! 


24.76 1 


27.855 ! 


30.95 


34.045 


37.08 1 


40.205 


46.425 




52 


15.437 


19.296 1 


22.512 1 


25.728 1 


28.944 ! 


32.16 


35.376 


38.592 I 


41.808 


48.240 




53 


16.041 ' 


20.052 ; 


23.394 


26.736 1 


30.078 


33.42 


36.762 


40.104 


43.446 


50.13 




54 


16.656 : 


20.82 


24.29 ' 


27.76 ' 


31.23 1 


34.7 


38.17 1 


41.64 1 


45.11 


52.05 




55 


\T.27b 


21.594 


25.193 


28.792 


32.391 


35.99 


39.589 


43.188 j 


46.787 


53.985 




56 


17.909 


22.386 


26.117 


29.848 


33.579 1 


37.31 


41.041 


44.772 1 48.503 1 


55.965 




57 


18.557 ' 


23.296 1 


27.062 


30.928 : 


34.794 


38.66 


42.526 


46.392 j 50.258 ] 


57.99 




58 I 


19.214 


24.018 1 


28.021 1 


32.024 ; 


36.027 i 


40.03 


44.033 


48.036 52.039 1 


60.045 




59 ■ 


19.902 ; 


24.852 


28.994 ; 


33.136 


37.278 1 


41.42 


45.562 , 


49.704 53.846 


62.13 




60 1 1 


20.558 1 


25.698 


29.981 j 


34.264 


38.547 1 


42.83 


47.113 1 


51.396 55.679 | 


64.245 





JO HIGH-SPEED ENGINE DATA. 

The formula shows that for a given horse-power (= H) and 
size (= S X D) of engine, the value of -^ may be made to vary 

w 

with the three factors p^, -- and w"^ or N. 

The principal quantities which effect the value oi pm are the 

AB 
initial pressure /^ = ^y and the cut-off-—^; the most important 

considerations governing the choice of these two quantities will 

be given a Httle later. The value -^ cannot become smaller than 

a certain value -— - prescribed by the strength and stiffness of the 

/x 

reciprocating parts. The following limits employed in practice 
for high speed engines may be of service : 

w varies ordinarily from 600 to lOOO ft. per m. 

W 

-^ " *' " 2 to 6 lbs. 

A 

— =" " " 0.8 to 2. 

5 
The larger values of w* and the smaller values of-— are usually 

employed for large engines. In using the following tables we 

TT 

first assume a diameter and with our given value of — (= I. 

pm 

H. P. for each pound of mean effective pressure per □" of piston 

area) we find from table IX the corresponding piston speed, or we 

TT 

assume a given piston speed and find from it and — the corre- 

pm 

S 
sponding diameter. If we now assume a given ratio of- we 

W 
can get from Table X the number of revolutions A^and -— -, or if 

A 



PRELIMINARY ESTIMATE OF DIMENSIONS. /I 

we assume — we can, from the same table, ejet S and N. It should 

be noticed that within certain limits we can diminish TV and yet 
increase vS without disturbing other established quantities. 



Preliminary Estimate of Dimensions of Reciprocating Parts. 

Calculate the weight of hollow piston head, the proportions 
being assumed like those given in Reuleaux's Constructeur, Fig. 
798, p. 746, but with an additional plate at the lower end so that 
the head will be closed at both ends. 

Also estimate the weight of piston rod from the table on p. 75 1 
of Reuleaux's Constructeur, remembering that the length of pis- 
ton rod is greater than the stroke by, length of piston head, length 
of stuffing box* length of cylinder cover and amount that enters 
cross-head. The weight of the cross-head in the present design 
is small, and in this rough estimate may be taken at 35 lbs. 
The weight of the connecting rod may also be got at roughly 
by. supposing it to be of uniform rectangular cross-section 
throughout, its dangerous cross-section being calculated from 
the formulas given at the end of Klein's Elements of Machine 
Design. Assume the ratio of depth of connecting rod to its width 

h . 

to be — = 2. Adding together the weight of piston head, piston 

rod and the connecting rod, and then dividing the sum by 

W 
the area of the piston we get the quantity — - which must be 

equal or less than the quantity -- prescribed, by the chosen di- 

mensions and by the speeds or value — °. 



72 



TABLE OF WEIGHTS OF RECIPROCATING FARTS. 



TABLE X. 

W Fo W 
Values of — - -^ — = -— = 
A A Fo 



.0000 1 42 ^iV'^ 



W F 

To FIND — r FROM TABLE MULTIPLY BY ASSUMED VALUE OF ^ • 
A A 

w 

— = weight of reciprocating parts per Q" of piston. 

S — length of stroke in inches. N — revolution per minute. 

p 

-^ = average accelerating force per □" of piston. 

Quantities bracketed in table are length of stroke = S^ 

(2NS\ 



§.sg 

II! 


Revolutions per minute = TV. 


100 


150 


200 


250 


300 


350 


400 


450 


1 i 

500 ! 600 


600 


(36.0) 
.1955 


(24.0) 
.1303 


(18.0) 
.0978 


'"o^i. 


(12.0) 
.0652 


(10.29) 
.0559 


(9.0) 
.0489 


(8.0) 
.0435 


(7.2) 
.0391 


1 (6.0) 
1 .0326 


650 


(39.0) 
.1788 


(26.0) 
.1186 


(19.5) 
.0889 


(15.6) 
.0712 


(13.0) 
.0593 


(11.14) 
.0508 


(9.75) 
.0444 


(8.67; 
.0395 


(7.8) 
.0356 


(6.5) 
.0297 


700 


(42.0) 
.1676 


(28.0) 
.1118 


(21.0) 
.0838 


(16.8) 
.0670 


(14.0) 
.0559 


(12.0) 
.0479 


(10.5) 
.0419 


(9.33) 
.0372 


(8.4) 
.0335 


(7.0) 


725 


(43.5) 
.1618 


(29.0) 
.1078 


(21.75) 
.0809 


(17.4) 
.0647 


(14.5) 
.0539 


(12.43) 
.0462 


(10.88) 
.0405 


(9.66) 
.0360 


(8.7) 
.0324 


(7.25) 
.0270 


750 


(45.0) 
.1564 


(300) 
.1067 


(22.5) 
.0782 


(18.0) 
.0626 


(15.0) 
.0521 


(12.76) 
.0447 


(11.25) 
.0391 


(10.0) 
.0348 


(9.0) 
.0313 


(7.5) 
.0261 


775 


(46.5) 
.1510 


(31.0) 
.1010 


(23.25) 
.0767 


(18.6) 
.0606 


(15.5) 
.0505 


(13.31) 
.0433 


(11.63) 
.0379 


(10.33) 
.0337 


(9.3) 
.0301 


C7.75) 
.0252 


800 


(48.0) 
.1466 


(32.0) 
.0981 


(24.0) 
.0733 


(19.2) 
.0587 


(16.0) 
.0489 


(13.71) 
.0419 


(12.0) 
.0369 


(11.66) 
.0326 


(9.6) 
.0293 


(8.0) 
.0244 


825 


(49.50) 
.1422 


(33.0) 
.0948 


(24.75) 
.0710 


(19.8) 
.0570 


(16.5) 
.0474 


(14.4) 
.0406 


(12.38) 
.0356 


(11.0) 
.0316 


(9.9) 
.0284 


(8.25) 
.0237 


850 


(51.0) 
.1380 


(34.0) 
.0920 


(25.5) 
.0690 


(20.4) 
.0553 


(17.0) 
.0460 


(14.57) 
.0394 


(12.75) 
.0345 


(11.33) 
.0307 


(10.2) 
.0276 


(8.5) 
.0231 


875 


(52.5) 
.1341 


(35.0) 
.0893 


(26.25) 
.0670 


(21.0) 
.0536 


(17.5) 
.0447 


(14.71) 
.0383 


(13.13) 
.0335 


(11.66) 
.0298 


(10.5) 
.0268 


(8.75) 
.0223 


900 


(54.0) 
.1303 


(36.0) 
.0868 


(27.0) 
.0652 


(21.6) 
.0521 


(18.0) 
.0434 


(15 43) 
.0377 


(13.5) 
.0326 


(12.0) 
.0290 


(10.8) 
.0261 


(9.0) 
.0217 


950 


(57.0) 
.1235 


(38.0) 
.0823 


(28.5) 
.0618 


(22.8) 
.0506 


(19.0) 
.0412 


(16.29) 
.0353 


(14.25) 
.0309 


(12.66) 
.0274 


(11.4) 
.0247 


(9.5) 
.0211 


1000 


(600) 
.1172 


(40.0) 
.0782 


(30.0) 
.0587 


(24.0) 
.0469 


(20.0) 
.0381 


(17.14) 
.0335 


(15.0) 
.0293 


(13.33) 
.0261 


(12.0) 
.0235 


(10 0) 
.0196 


1050 


(63.0) 
.1117 


(42.0) 
.0745 


(31.5) 
.0553 


(25.2) 
.0447 


(21.0) 
.0373 


(18.0) 
.0319 


(15.75) 
.0280 


(14.0) 
.0241 


(12.6) 
.0224 


(10.5) 
.0186 


1100 


(66.0) 
.1066 


(44.0) 
.0711 


(33.0) 
.0533 


(26.4) 
.0427 


(22.0) 
.0356 


(18.86) 
.0305 


(16.5) 
.0267 


(14.66) 
.0237 


(13.2) 
.0213 


(11.0) 
.0182 


1150 


(69.0) 
.1020 


(46.0) 
.0680 


(34.5) 
.0510 


(27.6) 
.0418 


(23.0) 
.0340 


(19.71) 
.0292 


(17.25) 
.0254 


(15.33) 
.0221 


(13.8) 

.0204 


(11.5) 
.0174 


1200 


(72.0) 
.0978 


(48.0) 
.0652 


(36.0) 
.0489 


(28.8) 
.0391 


(24.0) 
.0326 


(20.57) 
.0279 


(18.0) 
.0244 


(16.0) 
.0217 


(14.4) 
.0196 


(12.0) 
.0163 



DETERMINATION OF WEIGHT OF FLY-WHEEL RIM. 73 

VI. 

Determination of Weight of Fly-wheel Rim.* 
Let Wo represent the mass of the fly-wheel, Wo its weight, V^ 

F, + F, . 
its maximum, K its minimum, and V = — its average 

2 

speed of rim in feet per second. Then will maximum variation 
of energy equal 

— {K' - K') = ^° (F- + K){K - K) (5 1) 

2 ^ 

the coefficient of unsteadiness is 



hence variation of energy = 

/moV^=/ -° V^ = .00008 5 / Do'N^zVo ( 5 3) 

Dq being diameter of rim in feet. N = Revolutions per minute 
From the phase of greatest variation of tangential pressuse dia- 
grams we get Mm and MO N then if A = area of piston in Q'' 
and R' = length of crank in feet we have variation of energy = 



MmX A X — ^ X 2nR' z:^ .000085 f D^N^w^ (54) 
360 



Ax MONxMmX R' .. 

^,^,05 ^-^^^ (55) 

According to Der Taschenbuch des Ingenieurs the coefficient of 

unsteamess/= — — where V= varies as follows, 

F 2 



* The figure and formulas of this article were taken from Rigg's Treatise 
on the Steam Engine. 



' to 

20 


I 

30 


25 


35 


i- to 

30 


I 
40 


i to 

40 


I 
60 



74 DETERMINATION OF WEIGHT OF FLY-WHEEL RIM. 

For machines which will permit a very uneven 

motion, hammers, etc., /■= — 

5 
For machines which permit some irregularity, 

pumps, shearing machines, etc., / = 

For machines which require approximation to 
uniform speed, as in flour mills, / = 

For machines with tolerably uniform speed as 
weaving and paper making, / = 

For cotton-spinning machinery requiring very 
uniform speed, / ^ 

For the spinning machinery of very high yarn 

numbers,* / = — 

100 

The areas (see figure) included between mean tangential 
pressure circle and that portion of the tangential pressure curve 
lying outside of the mean tangential pressure circle we will call 
phases of excess of power ; the areas included between the mean 
circle and that portion of tangential pressure curve within the 
circle of average resistance we will call the excess of resistance. 
These phases may be numbered as in figure by Roman numerals, 
and the angles at the center which they subtend, in degrees, 
The average excess or deficiency of pressure of any phase may 
be found tentatively by drawing arcs subtending the angle of 
each phase as in figure, the area included between these arcs and 
the mean tangential circle being equal to the areas of their cor- 
responding phases. 

7 ■ .y. . • Excess of power or resistance during any phase 
' Total ipower exerted during a revolution 

Mm MON 

= X = coefficient of variation of energy. 

Aa 360 

_ __ _ 

* For electric lighting, / = — > ^ "^ fPds 



DETERMINATION OF WEIGHT OF FLY-WHEEL RIM. 



75 




^^S- 33- 



/ = coefficient of unsteadiness = 



K 



F 



"^ when V= Z^+Z^ 



= mean velocity of rim. / = — in present case. 

A =z area of piston in □''. 

R = length of crank in inches. 

Mm = excess of pressure in lbs. (see figure). 

MON = angle (in degrees) of phase of greatest variation. 

A 
N 
H' 



V 



W. 



= diameter of fly-wheel in feet. 

= revolutions per minute. 

= indicated horse-power of engine when the greatest 

variations of power or resistance occur. 
= velocity of rim per second ^ 80 feet. Assume = 64 

feet in present case. 
= weight of fly-wheel rim. (It may be diminished 8 % 

for arms and boss.) 



76 DETERMINATION OF WEIGHT OF FLY-WHEEL RIM. 



,„ AR Mm MON , ^. 

"^='7-1- f^.^. (56) 

or 

kH' 
Wo = 388000000 j^jj;^ (57) 

If in the figure on the preceding page Mm = 2.9 Nn = 1.9, 
N'm = 1.9, M'n = 1.6 and Aa ^ p^ = 3.1, the areas of the 
various phases will be measured by the following products 

I Excess of Power = 2.9 X 77° = 223.3 

II " " Resistance = 1.9 X 111° = 210.9 

III " " Power = 1.9 X 75° = 142.5 

IV " " Resistance = 1.6 X 97° = 155.2 

Excess of Power = 365.8 Excess of Resistance = 366.1 

Had the work been perfectly accurate the two results would have 
been exactly equal. It is evident that in the present example I 
is the phase of greatest variation, consequently its average ordi- 
nate Mm and angle MON should be substituted in formula for 
weight of fly-wheel rim. The ratio which the measure 223.3 of 
phase I bears to the measure Aa X 360 = 3.1 X 360 =1116 

22^ ■s 

of the total work done in one revolution is — ^ z= 0.22 == k. 

1116 

Find from the three tangential pressure diagrams for //", H^, 11^, 
the phase of greatest variation and substitute in formula for fly- 
wheel. 

In a manner similar to that detailed on preceding page we can 
find the phase of greatest variation from the tangential pressure 
diagrams drawn on the rectified crank-pin circle. Instead of 
multiplying the average pressure of each phase by the size of the 
phase expressed in degrees, we multiply by the distance between 
the two intersections of each phase with the mean tangential 
pressure line. 



BALANCING RECIPROCATING PARTS BY COUNTER-WEIGHT. // 



VII. 

Balancing the Reciprocating Parts by Counter-weight 
ON Crank. 

We have already seen that only a portion P' of the effective 
steam pressure P is transmitted to the crank pin during the first 
portion of the stroke, the remainder (P — P') being engaged in 
accelerating the reciprocating parts. The absorption of pressure 
by the moving parts disturbs the statical equilibrium of the forces 
acting on the engine bed BCD A, The steam pressure P acting 
on the rigid frame at A being greater than F the horizontal force 
at B, the tendency is to shift the engine bed in the direction CD 




_J LJ 



Fig. 34- 

of the greater force P, the shifting force being equal to P — P' . 
Similar considerations will show that as the piston approaches the 
end of its stroke this shifting force changes in direction and tends 
to slide the engine bed on its foundations in the opposite direc- 
tion DC. In order that this effect may be wholly or partially 
neutralized by other means than the employment of numerous 
bolts for holding the engine bed to its foundations, a counter- 
weight m or its equivalent is attached to the crank as shown in 
the figure. The horizontal component P — P'' of the centrifugal 
force xFq developed by this mass m when revolving in the crank- 
pin circle with velocity Fwill then be transmitted to the point 
B of the engine bed and together with P' will make a force equal 
to Pand will balance the steam pressure P acting at the other 
end A of the frame. 
6 



yd) BALANCING RECIPROCATING PARTS BY COUNTER-WEIGHT. 

The horizontal component of the centrifugal force is easily ob- 
tained from the expression 

X p^ cos io ^ m ~ cos o) = — - cos co {^S) 

K g K 

where m represents the mass engaged in balancing the reciproca- 
ting parts reduced to a point on the crank-pin circle diametrically 
opposite to the crank pin, v =z velocity of crank pin, R = radius 
of crank, co = crank angle, we have already shown that the force 
F accelerating or retarding the reciprocating parts is given by the 
expression 

F= — ;^{ cos ^'^ ^ Y cos 2co\ (59) 

where W represents the weight of the reciprocating parts. 
The frame will be in statical equilibrium and will have no ten- 
dency whatever to shift on its foundations 

to i>^ W v^r . R 

— -- cos CO z=i -- 

£^R g R 



when - -^ cos to z=i — "5"l cos <w -|- - cos 20 \ (60) 



that is when 



to == ^(cos CO ^ - cos 2co\. (61 



cos CO 



It is evident from this equation that equilibrium cannot exist for 
all values of co unless the counter-weight tD also varies with the 
crank angle ca. This it is of course not practicable to do, hence 
there will always be a shifting force acting on frame equal in 
amount to 

WvU , R \ vsv^ 

z=r I cos CO ^ — COS 2CO ] -— cos CO 

g R^ L y gR 

= FA COS CO -\- — cos 2CO — X cos COj 

= Fo\ (i — ;i^)cos CO -\- - COS 2co (62) 

X being equal to -— . 



BALANCING RECIPROCATING PARTS BY COUNTER-WEIGHT. 79 

The maximum shifting effort when reciprocating parts are not 
balanced by counter- weight corresponds to <^ = o and x = o; 
hence maximum shifting force equals 

R 



(-!) 



and when the reciprocating parts are balanced the maximum 
shifting force corresponds again to <x> = o and is equal to 

R 



".(--+!) 



When ,r = unity, that is when counter-weight m = W =^ weight 
of reciprocating parts, the ratio of these two maximum shifting 
forces is 

R / R\ 

It will generally suffice to make ;tr := .5 to .8 ; the unbalanced 
portion being resisted by the foundations. 

Calculate the value of the vertical components of the mass m. 

Instead of employing the counter-weight as shown in the pre- 
ceding figure it is customary to employ a crank disk of the fol- 
lowing shape. 

The crank proper is balanced by a counter-weight of similar 
shape shown in dotted lines, there remain therefore only the two 
portions ABODE and A' B' C D' E having the depth w which can 
be utihzed as counter-weight for balancing the reciprocating parts, 
the dotted lines A' F' G' E' and AFGE representing the counter- 
balance for part (if any) of the weight of connecting rod, which 
part is supposed to be concentrated at crank pin. It is evident from 
the figure that the center of gravity c of the portions ABCDEGf 
and A'B' C D' E'F' does not fall upon the crank-pin circle, conse- 
quently this deficiency of radius must be made good by increas- 
ing the masses ABCDEGF2.nd A' B' C D' E' G' F' . Let the sum 



8o 



BALANCING RECIPROCATING PARTS BY COUNTER-WEIGHT. 



of these larger masses be represented by M\ its radius by r, then 
we must have 

M'(v'y _ m'v"" 
r 



R 



and smce — z= -— we have M' = —m. 
V R r 



(63) 



The value of r can be found experimentally by suspending the 
figure ABCDEGF {rom two of its corners and determining the 




1_. 

1 
1 

: i, 

1 




\ 










J u* 


1 
t 

6— u.-> 







Fig. 35- 



intersection of the two lines of suspension, or it may be balanced 
over a knife edge in two different positions. M' can then be 
computed. 

The dimensions of the simple crank can be calculated from the 
formula given by Reuleaux. 

To find room on the crank disk for the large mass M' of coun- 
terweight, the outside circumference of crank disk, must be taken 
considerably larger than crank-pin circle. By casting hollows in 
the disk and filling them with lead, all the counter- weight desired 
can be placed on crank disk. When double cranks or disks are 
employed the disposal of counter-weight becomes an easy matter. 



INFLUENCE OF NEGLECTED RESISTANCES. 8 1 



VIII. 

Effect of Frictional Resistances, of the Weight of the 

Rod AND Exact Values of the Forces of Inertia, on the 

Rotative Effort, on the Pin Pressures and on 

the Force Shaking the Engine Bed. 

The effect of these three influences on the rotative effort and 
energy of the fly-wheel has been very fully considered by Prof 
D. S. Jacobus* in two papers, published in Vol. XI, Transac- 
tion of American Society of Mechanical Engineers. 

In these papers very accurate formulas were established which 
took account of these three influences. They were applied to 
four different engines representing a wide range of practice. The 
exact diagrams constructed for these engines showed that the 
fluctuation of energy, AE, of the fly-wheel due to the variation of 
the rotative effort differed by an insignificant amount from the 
approximate diagrams found by supposing all the reciprocating 
parts (inclusive of rod) concentrated on piston. A similar result 
was obtained for the crank-pin pressures, the nearly exact ones 
differing but slightly in direction and intensity from those found 
by assuming the mass of the rod to be all concentrated at wrist- 
pin. The following tables are given by Prof Jacobus and furnish 
the data of the four engines and the results of the numerical 
computations. 

* Prof. Jacobus credits Prof. J. B. Webb with valuable assistance in the 
preparation of these two papers. 



82 



DIMENSIONS AND SPEED OF ENGINES. 



TABLE XL 

Dimensions and Speed of Engines to which the Formulas 
have been applied. 



Revolutions of crank shaft per minute . . 

Length of stroke, in inches 

Diameter of cylinder, in inches 

Length of connecting-rod, in inches . . . 
I Distance from the wrist-pin to the center of 
i gravity of rod, in inches 

Distance from the center of the crank shaft 
to the line of travel of the wrist-pin, in 
inches 

Principal radius of gyration, in inches . . . 

Weight of piston, piston-rod, and cross-head, 
in pounds 

Weight of connecting-rod, in pounds . . . 

Indicated horse-power 



Class of Engine. 



a; 






300 

12 
10 
36 

2C.I5 



O 
15.00 



II 



biO • 

11 
■'J 



250 

24 

92 

55 



o 

34-I 



90. |474 
307 

345=* 



70. 

57 



III 



X3 

tn ^- 
o ^ 



II 

b|OoU 
0(5 
h4 



60 
60 

26 yi 
150 

78 



o 

48 

1300 
1200 

346 



IV 



3 
o 

c 



120 



II 
5^ 

3.22^ 



0.57? 
2.077? 

100 

50 
66 



For one cylinder. 



NUMERICAL APPLICATION TO STANDARD CLASSES OF ENGINES. 83 



TABLE XII. 

Results of Numerical Applications to Several Standard 
Classes of Engines. 



So 

c 
W 

"o 

en 

tn 
ctf 

D 
I 


Conditions assumed. 


fPds 


Pressure acting on 
crank pin 


Exact, 


Approx- 
imate. 


Not in- 
cluding 
acceler- 
ating 
forces. 


Exact. 


Approx- 
imate. 


Not in- 
cluding 
acceler- 
ating 
forces. 


Not including the effects 
of friction and weight 


.184 


.174 


.238 


58.1 


57.3 


83.0 


I 


Not including friction but 
including weight . . 


.180 






58.0 
56.8 




. . . 






I 


Including both friction 
and weight , . . . 


•171 
.181 










.172 


.275 




i 


II 


Not including friction and 
weight 


56.8 


55-5 


108.0 


III 

IV 


Not including friction and 
weight 


.142 


•131 

.147 


.185 

.247 


76.1 


75-5 


89.4 


Not including friction and 
weight 


.160 


61.0 


60.8 


75-5 



The term approximate in these tables refers to the assumption 
that the whole mass of the rod is concentrated at wrist-pin. The 

AE . . 

expresssion zr-rrr is the ratio of the periodical excess or deficiency 
J Fas 

of energy AE, to the whole energy exerted per revolution, f Pds. 
The particular crank-pin pressure given in table is the maximum 
one and is in pounds per square inch of piston area. 

It is evident from these tables that for most practical work the 
aforesaid approximation, used in the preceding sections, is suffi- 
ciently accurate. The remainder of this section may therefore 
be omitted by those who do not care to examine the refinenients 
of this subject. 



84 ORDER OF DISCUSSION. 

We have not reproduced here any of Prof. Jacobus* formulas 
for the components of the forces acting on the pins, though they 
are easily applied, and, after the constants have been computed 
and introduced, lead quickly to the desired results. We have 
preferred graphical determinations because they are simple, 
and naturally accompany designing. We will therefore follow 
graphical methods in finding exact values for each of the three 
influences under consideration. 

The order in which we will take them up is : 

a. The friction between a pin and its bearing. 

b. The direction of internal stress in a rod, when friction 
is taken into account, but gravity and inertia neglected. 

c. The direction of internal stress in a rod, when friction 
and some other force, say, resultant of gravity and in- 
ertia-resistance, are considered. 

d. Determination of the exact accelerating force of the 
rod corresponding to its motion in the slider-crank 
chain. 

e. Combination of weight of rod with its force of inertia. 

f. The components of this total force at wrist- and crank- 
pin, when friction is neglected. 

g. The components of this force at these pins, when fric- 
tion is considered. 

h. Determination of the force shaking the engine bed. 
i. Diagrams of shaking forces with different degrees of 

counterweighting. 
j. Diagrams of pressures at crank- and at wrist-pin. 

A. 

The Friction Between a Pin and its Bearing. 

The frictional resistances of motion and rest differ by quantities 
that are directly proportional to the coefficients of friction for 
motion and rest, respectively. By using the coefficient of friction 
for motion, we may treat the body as if it were at rest and yet 



THE FRICTION BETWEEN A PIN AND ITS BEARINGS. 



85 



determine the direction and intensity of the forces as they exist 
under running conditions. 

Let us suppose the pin to be at rest, without tendency to move 
in either direction, and that the resultant pressure P^ between the 
pin and its bearing acts in the direction c a e f, Fig. 36. Then 
the pin can be brought to the eve of, say, left-handed rotation, in 
one of two ways, either by the action of an additional, single, 
force Q, with lever arm c /, or by the action of some couple M 
having left-hand rotation. 




Fig. 36. 

In the first case there will be an infinite number of solutions 
depending upon the location and direction of the turning force 
Q. The moment F'r of the friction of the bearing will balance 
the moment Q X ^/ of the new force and the point of application 
of the resultant pressure between pin and bearing will be shifted 
from a to b, sufficient play being supposed to exist between the 
pin and its bearing to permit this to take place. The point b is 
where the resultant P(=eg) of the forces 2(= ^^) and Po(=e f) 
cuts the area of contact. Evidently the new resultant Pis dif- 
ferent in intensity and direction from the original pressure P„. 
The pin is kept stationary by -(- Pand — P\ turning of the pin 
is prevented because 

Q X cr= P X cd= Pf/ = F'r = <pNr. 



S6 THE FRICTION BETWEEN A PIN AND ITS BEARING. 

The resultant pressure P between bearing and pin is therefore 
necessarily a tangent to the friction circle whose radius p' = (pr 
when P is taken to equal N. This is permissible because the 
angle of friction is so small, that we may regard its sine and tan- 
gent as equal. 

The force P for this first case can also be constructed by plac- 
ing at center c two equal and opposite forces Q^ z= Q^ =: Q. 
Then Q and Q^ will form a couple which, combined with the 
original force P^, will shift the latter parallel to itself through a 

p 
distance -— />'; now combining P^ in this new position with the 
-^ 

force Q^ remaining at the center we get the same force P as 
before. 

In the second case the pin is brought to the eve of left-hand 
rotation by the action of some left-hand couple J/ which may 
act anywhere in the plane. This case does not often arise in 
practice, and is chiefly useful in demonstrations involving equiv- 
alence of forces. The effect of this couple is to shift the force 
Po parallel to itself to a position in which it is tangent to the so- 
called " friction circle " having radius p' = (pr. The point of 
contact for this case and the direction and intensity of its force 
P are all different from those obtained for the first case. As the 
couple M may act anywhere in the plane, let us suppose it re- 
placed by an equivalent couple Qo X cl whose two equal and 
opposite forces Q^ act respectively at the center C and along the 
line of turning force Q mentioned in the first case. Then it is 
easy to show that this force Q^ is different from Q. In Fig. 36, 
we have given, the angle ;-, the radii r and p' and the lever arm cl. 
Then 

^ cd cd sin r p' sin r, . P sin r 

sm o = ^=r = — =^- = - — =— ^ and — - = 1 — • 

ce cl cl Po sin (r— 07 

but QyTd^Py^p' 

__ Q P sin(r — o) ,^ . 

and QoXcl=P,y. p' then q; =" p^ = smy ' ^ "^^ 



INTERNAL STRESS IN A ROD ACTED UPON BY TWO PINS. 87 

Q^ = Q = fL^!^. (65) 

p. p d d ^ ^' 

As each piece is subjected to at least two forces, the driving 
force and the resistance, there will always be a force Q available 
for bringing the pin or shaft to the eve of turning. We will 
therefore assume the first case as the common one. It should be 
noticed that the resultant action of bearing on pin is tangent to 
that side of friction circle which will give this resultant a compo- 
nent opposed to the rotation of the pin or shaft. In like manner 
the action of pin on bearing is such a tangent to the friction circle, 
that it has a component opposed to the bearing's (relative) rota- 
tion. This is because friction is a hindrance to motion. In Fig. 
36, these equal and opposite actions are indicated by arrows and 
the letters P on B and B on P. 



B. 

The Direction of Internal Stress in a Rod acted upon 

BY Two Pins. 

There are four cases under this head, which are due to the two 
kinds of relative motion possible between a pin and its bearing, 
and to the two kinds of stress, tension or compression, to which 
the rod may be subjected. The four figures* on next page illus- 
trate these four cases. 

We will suppose the link AB, Fig. 37, in all cases to carry the 
eyes or bearings into which fit the pins belonging to the cranks 
MA and NB. The large arrows about the center iVor i)/ indi- 
cate which crank is the driver and the small arrows about the 
pin centers A and B indicate whether the rotation of rod AB to 
crank MA or A^B is right- or left-handed. The arrows on the 
rod AB itself indicate whether it is in tension or compression. 

* These figures were taken from Hermann-Smith's Graphical Statics 
of Mechanisms. 



88 



INTERNAL STRESS IN A ROD ACTED UPON BY TWO PINS. 



In cases I and II the relative motions of the pins A and B to 
their bearings in the rod AB are alike, being both right-hand 
rotations, but the internal stress is different in the two cases, being 
tension in case I and compression in case II. 

In cases III and IV the relative motions of pins B to their 
bearings in the rod are still right-handed, but those of the pins A 
to their bearings are now left-handed. The inner stresses are also 
different in these two cases, tension in case III and compression 
in case IV. 




Fig. 37. 



When there is no friction the inner stress acts along the center 
line of rod in each case. 

When there is friction, the direction of the inner stress is 
different in each of these four cases. 

In Figs. 38-41 sections are taken of both pin and its bearing 
near the point of contact and arrows added to indicate the rela- 
tive motion of rubbing surfaces. 

Now that there is friction, we do not at first know the point of 
application of the resultant of the forces acting at each surface of 



INTERNAL STRESS IN A ROD ACTED UPON BY TWO PINS. 



89 



contact; but we do know that the action ap, Fig. 38, of pin A 
on bearing must have a component opposed to the (relative) 
motion of the bearing. This action ap prolonged backward will 
therefore be tangent to lower portion of friction circle cde. In 




Fig. 38 



like manner the action of pin B on its bearing when prolonged 
backward will also be tangent to its friction circle fgh, but at the 
upper portion of the circle. Neither ap nor bp represent the two 
directions of the forces exerted by the pins, for friction has 
changed the intensities of these forces and shifted their points of 
application to a! and b' . As there are but two forces acting on 
the rod, it can only be in equilibrium when the two forces are 
directly opposite and equal. This condition determines the 




Fig- 39. 



direction of the inner stress ; it must be tangent to each of the 
friction circles of A and B as in the figure, the pin A acting at a' 
and pin B at point of application b' . 

In the case shown in Fig. 39 the rotation of the pins is still 
right-handed, but the rod is now subject to compression and con- 



90 



INTERNAL STRESS IN A ROD ACTED UPON BY TWO PINS. 



tact takes places at a and b on sides of the pins opposite to those 
of Fig. 38. Similar reasoning shows that in this case the direc- 
tion of the inner stress must be along the line 2eh2, 




Fig. 40. 



In Fig. 40 pin B keeps its right-hand rotation, but pin A has 
left-hand rotation, the rod being in tension. The figure indicates 
the condition of equilibrium and direction 3^3 of the inner stress. 




Fig. 41. 



In this last case, Fig. 41, pin B has still right-hand rotation 
and A left-hand rotation. Compression however now exists in 
the rod and this brings the contacts a and b different from the 
preceding case, namely, on the opposite sides of the pins A and B, 
The tangent ch on line 44 is the direction of the inner stress. 

These four cases have furnished four lines of stress coinciding 
with the four possible tangents, ii, 22, 33 and 44 to the friction 
circles. This would also have been the case, if the rod had 
carried the pins instead of the eyes or bearings. 



DIRECTION OF INTERNAL STRESS IN ROD. 9I 



c. 

Direction of Internal Stress in Rod when its Acceler- 
ating Force is known. 

The known accelerating force KG, Fig. 42, is the resultant of 
the unknown forces acting at the crank-pin C and wrist-pin W. 
When there is no friction we will designate these forces by Pc and 
P^ respectively and their directions must be assumed to pass 
through their respective centers C and W. When there is fric- 
tion the forces will be designated by Pc/ and P.^^/ and they must 
then be tangent to their respective friction circles. In either case, 
unless some additional condition is given with reference to one 
or the other of these forces, the value of Pc or Pc/, Pw or P-w/ is 
indeterminate for there are evidently in each of these cases an 
infinite number of pairs of forces which can produce the given 
accelerating force KG. 

The forces P^/ and /^ (or Pc and P^) evidently produce 
both acceleration PH (= KG) and whatever internal stress may 
exist. This latter will vary according to the pair of forces 
assumed to produce the acceleration and also according to the 
points of application assumed for the members of this pair. For 
instance, let us assume that PF ^.nd P^ represent the directions 
of the pair of forces Pc/ and P^^y, then any points F' , /% 5 and 
E\ E, T'm the direction of these forces may be assumed as the 
points of application of the forces, and the corresponding direc- 
tions of the internal stress will be ST, FE, F'E . The intensi- 
ties of these internal stresses along either of these directions will 
depend entirely upon the acceleration components of the forces 
P^f and Pcf. These components acting at the assumed points of 
application 5, 7" or F, E or F\ E' have only one condition to 
fulfill, namely, to produce the acceleration PH. As this can be 
accomplished in an infinite number of ways, at the same pair of 
points of application, it is evident that those other components of 
P^and Pcfy the internal stresses, have also an infinite number of 



92 



DIRECTION OF INTERNAL STRESS IN ROD. 



values. But when the directions of these acceleration compo- 
nents have once been decided upon, their intensities and those of 
their fellow components (the internal stresses), can be found by 
resolving each of the forces P^f and Pc/ along the directions 
assumed for their components. According to D'Alembert's prin- 
ciple* the internal stresses thus found must balance each other. 




Fig. 42. 



To illustrate what has been said, let us suppose the points of 
application chosen to be 5 and T, then TS will be the line of 
internal stress ; now assuming the directions of the acceleration 
components of P^f and Pc/ to be SZ and and ZZ, then making 



* According to D'Alembert's principle those components of the external 
forces which are not engaged in producing acceleration are in equilibrium. 



FORCE NEEDED TO ACCELERATE ROD. 93 

Sm =z PM = P^af and Tn = MH = Fc/, we get Sd and Tq as 
the acceleration components and Sd, Ta the components produc- 
ing the internal stresses in the rod along the line TS. Measure- 
ment will show that Sd = Ta. In like manner we might have 
assumed the acceleration components Sd' and T^' both parallel 
to the total acceleration KG ; then resolving P^jr and P^/- we get 
Sd' -\- Tq' = KG and internal stress Sd' = internal stress TV. 
If F and £ had been chosen as the points of application, FB 
would have been line of internal stress. Then if FV and £V 
had been chosen as direction of acceleration components, 
Fe = Eh would represent the equal internal stresses. It is evi- 
dent that there is great freedom in the choice of the direction of 
the internal forces and later on we will indicate how this may be 
so chosen as to lead very directly to the determination of the 
actual pressures exerted by the pins upon their bearings in the 
connecting rod. In the engine problem these pressures are not 
indeterminate as they are here, for there the known effective 
steam pressure and the direction of the reaction of the guides, 
together furnish another condition, which is all that is necessary 
to completeley determine P^/. 

D. 

Exact Determination of Force Needed to Accelerate Rod. 

Any motion of a body in a plane is equivalent to the rotation 
ol the body about its instantaneous axis. The rotation about 
this axis is, by a principle of Kinematics, equivalent to an equal 
rotation about any other, second, axis plus a translation, the latter 
motion being equal to that possessed by the second axis when 
rotating around the first, or instantaneous, axis. This equiva- 
lence of motion extends not only to the displacements, but also 
to the velocities and accelerations. The axis passing through 
the center of gravity will be taken as this second axis because it 
is the most convenient one, the centripetal accelerations about 
this axis completely neutralizing each other. 
7 




94 FORCE NEEDED TO ACCELERATE ROD. 

The proposition may also be stated in this better known form : 
The total accelerating force of a body, whose points move in 
parallel planes, is compounded of the force needed to give the 
whole mass an acceleration of translation equal to that of its 
center of gravity and of a moment (or couple) capable of impart- 
ing to the body an angular velocity and acceleration about its 
own center of gravity equal to that possessed by the body when 
rotating about the instantaneous axis. 

The combination of this force of translation with the moment 
or couple, gives a single resultant force that is equal and parallel 
to its component force of translation but has a location that is dif- 
ferent from this component, the effect of the couple being merely 
to shift the location of the force producing the translation, a couple 

. , f left-hand 1 .,.,.,, 

with < . , , , > rotation shiftinpr the force to the 
I right-hand J ^ 

in applying this rule the plane of forces must be viewed so that 
the force of translation is downward. 

The moment of the force or couple capable of producing the 
desired angular acceleration — , is, according to Mechanics, 

I -T- = m^^ — - (66) 

dt dt ^ ' 

where I is the principal, polar, moment of inertia, Itt the mass 
of the connecting rod and k the corresponding radius of gyration. 
This expression is perfectly general so far as the distribution 
of the mass of the rod is concerned. Any arrangement of this 
mass which will not alter its center of gravity and which pre- 
serves the same total moment of inertia I unchanged will require 
the same total accelerating force as the rod itself. The problem 
may consequently be greatly simplified for graphical purposes by 
supposing the mass of the rod concentrated at two points that 
are on opposite sides of the center of gravity and in line with it. 



FORCE NEEDED TO ACCELERATE ROD. 95 

In order that the equivalence of total accelerating force may- 
be maintained it is necessary that the following three conditions 
be fulfilled, 

m r=z m, + m„ (67) 

mji, = mji^ (68) 

m/'" = m,/^," + mA^ (69) 

and by combining these we get 

k' ^ k,k,, (70) 

where kj and k^ are the distances from the center of gravity of the 
two points at which the mass is supposed to be concentrated and 
nil, lltz the masses at these points. Equations {6y) and (68) keep 
the total mass and location of the center of gravity the same as 
before and therefore keep unchanged the accelerating, compo- 
nent, force due to the translation of the center of gravity. The 
fulfillment of equation (69) or (70) makes the turning moment of 
the rearranged, two-point, rod the same as that of the original 
rod with distributed mass. As the force and energy of this two- 
point rod are the same as in the original one, we can confine our 
determination of the total accelerating force to the simplified, 
two-point rod. 

The problem is now reduced to finding the acceleration of the 
center of gravity of the rod and that of each of its two points of 
concentration. The problem is therefore now a purely kinematic 
one and is a special case of the general problem of finding the 
acceleration of any point on the connecting rod of the slider- 
crank mechanism. 

In this general case of the mechanism and its motion, the accel- 
eration CO, Fig. 43, of the crank-pin center C is known and the 
acceleration IVw'^ {= Ow) of the wrist-pin center or slide can be 
obtained as in Fig. 16 and p. 43. The accelerations of two points 
of the rod will then be known and the center of acceleration G 
of the rod can be found.* The acceleration of any point of the 



* See Weisbach-Hermann's Machinery of Transmission, Vol. Ill, Section 
I. §21. 



96 



FORCE NEEDED TO ACCELERATE ROD. 



rod is directly proportional to the distance of this point from the 
center of acceleration G, and, for the instant in question, the 
acceleration of each point makes the same angle with its own 




Fie. 43. 

instantaneous radius of acceleration.* This solves the problem 
but not in a convenient way because the center of acceleration 
generally falls beyond the limits of the drawing. A method that 
is free from this objection and also simpler in other respects is 
found as follows : 

In Fig. 43 revolve the triangle CWG (having the rod CW as 
a base and acceleration-center C as a vertex) through an angle 
GCG' , equal to that made by the direction of acceleration of 
point C with its instantaneous radius of acceleration CG, Then 
will the revolved radii G'C, G'^\ G' H' and G'W be parallel to 
the directions of the accelerations of points Cy C^, //and W, as 
well as proportional to the magnitude CO, ^g" Hh" and Ww" of 



* See last paragraph on p. 45 and first one on p. 46 for one method of 
finding acceleration of any point of rod. 



FORCE NEEDED TO ACCELERATE ROD. 



97 



these accelerations. As the triangle COw is similar to the triangle 
CGWdXid. Chgw to the rod CH^W, the vectors OC, Oh, Og, and 
Ow drawn from 6? as a pole will represent in direction and mag- 
nitude the accelerations of the points C, H, (^ and W of the rod. 

The pole is the end of the acceleration CO of crank-pin and 
when Ow is found by method given in Fig. i6 and p. 43, the 
figure Cwgh can easily be made similar to the rod CW^H. 
Since H may be any point in the rod we have a simple method 
of finding its acceleration (without first finding the center of 
acceleration) provided the direction and intensity of the accelera- 
tion of two points of the rod are known.* 

We can now give two simple constructions for the total force 
accelerating the rod of the ordinary shder-crank, when crank 
has uniform rotation. 




Fig. 44.— First Construction, 

The mechanism OCIV, Fig. 44, is the ordinary slider crank 
in which the stroke of the slide W^ passes through the center O 
of the crank OC. The center of the rod is at ^, the center of 
gravity at G and GF= GD is the principal polar radius of gyration^ 



* This same simple method may also be deduced from a more general 
case. See Journal Franklin Institute for September and October, 1891. 



98 FORCE NEEDED TO ACCELERATE ROD. 

We first construct the acceleration of the sHde W by the 
method given on p. 43 and Fig. 16. 

Prolong the rod IVC till it intersects at B the perpendicular 
OB to the stroke OIV. With ^'asa center and CB as a radius 
describe a circle Btv. On the rod CW as a diameter describe 
another circle. Join the intersections of these circles by the 
chord tsv and prolong it, if necessary, till it cuts a Hne drawn 
through crank center and parallel to stroke. In Figs. 44 and 
45 this intersection is at w, and Ow is the desired acceleration of 
the slide, provided crank length CO represents the centripetal 
acceleration of the uniformly rotating crank-pin. 

(The construction would be exactly the same for the crossed 
slider crank of Fig. 46 if its crank had uniform rotation. This 
construction of the slide acceleration has no failing case even at 
the dead point.) 

Now join C and tc. Cw is a reduced image of the rod, making 
with it an angle equal to that made by each point's acceleration 
with its own radius of acceleration and point has the same 
positions relatively to the image Cw of the rod that the center of 
acceleration has relatively to the rod itself It follows from this 
that if we draw through points F, G and D of the rod, parallels 
Ff, Gg and Dd to the stroke, that the distances or vectors Of, 
Og and Od will represent in direction and intensity the accelera- 
tions of /^, G and D, respectively. The actual locations of their 
accelerations will, of course, be parallels through these points to 
those vectors. One-half of the mass of the rod is supposed to 
be concentrated at i^and the other half at D. 

Draw FV parallel to Of and D V parallel to Od. Through 
their intersection Fdraw^ parallel to Og, the line VK; it will be 
the actual location of the resultant force of inertia, and its inten- 
sity will equal the product of the whole mass of the rod by 
acceleration Og of the center of gravity. 

In Fig. 45 the mass m of the rod is divided into two unequal 
parts, one of which ttti is concentrated at W and the other vx^ at 



FORCE NEEDED TO ACCELERATE ROD. 



99 




Fig 45.— Second Construction. 



//, the relation between the points W, G and //being GWX GH 
= k^, where k is the principal polar radius of gyration. The 
two masses must be to each other as 



mi 



m,= GH\ GW. 



Construct as before the acceleration Ow of the slide and Cw 
the image of the rod. At G erect the perpendicular GE = k. 
Join £ with Wand draw EH 2X right angles to EW. This will 
evidently satisfy the condition GW X GH = k^. 

Through H and G draw Hh and Gg parallel to the stroke, 
then will the vectors Qh and Og represent in direction and inten- 
sity the accelerations of points H and G, respectively. 

Through //draw ///parallel to Oh ; it will cut at /the direc- 
tion WI of the acceleration of the other mass at W. Through 
/.draw, parallel to Og, the line IK\ it will be the actnal location 
of the resultant force of inertia of the rod and its intensity = 
whole mass of rod X acceleration Og of center of gravity. 

These two constructions give, of course, exactly the same 
result, the first being a little the easiest when the intersection V 
falls within the limits of the paper. The great advantage of the 
second is that the direction of W\?> given by the line of stroke, 
and that the intersection /will always fall within the limits of the 

LoFC. 



lOO 



FORCE NEEDED TO ACCELERATE ROD. 



drawing. Both constructions of the total force of inertia fail at 
the dead point, but this is of no consequence, because then the 
direction and location of the total acceleration of the rod is 
known, for it coincides with the stroke, and, as before, its inten- 
sity = the whole mass X acceleration Og" of the center of 
gravity.* 

This construction is only one of an infinite number of possible 
cases, each of which satisfies the three conditions involved in 
equations (67), (68) and (69) or (70). 




Fig. 46. 

It is evident that hi = /^^ is a particular case of the general 
one just given, and is illustrated in Fig. 44. 

Fig. 46 is meant to be the most general case of slider-crank,with 



* The nearer point //, Fig. 45, is to crank-pin C the more nearly does the 
point of intersection / preserve a constant position. In the rods of good high- 
speed-engines the locus of point / varies but little, the average value of 0/ 

being nearly equal to C// — L — -j ; where L =^ CW xs length of 



hi 



connecting rod and hx = GJV, Fig. 45, is distance of center of gravity G 
from wrist-pin IV. This could be made the basis of a close approximation, 
in which the point / is treated as a fixed point and I/C is drawn at once par- 
allel to 0£-, thus omitting the drawing of the lines Nh, hO and HI. 



FORCE NEEDED TO ACCELERATE ROD. lOI 

respect to the mechanism itself, to its motion and to the distribu- 
tion of the mass of the rod. The mechanism LCWis of the 
crossed sHder-crank type in which the stroke of slide W does 
not pass through center L of crank. The rotation of crank CL 
is not uniform as in the preceding cases, the acceleration of crank, 
pin C, taking (say) the direction and intensity CO instead of CL 
as before. The center of gravity of the mass is no longer in the 
center line CIV of thQ rod, but outside of it, at G. The point H 
is on the line WG as before, and again satisfies the condition 
GJV X GN =z k^^ where k is the principal polar radius of gyra- 
tion. iYis found by the same construction, namely, we erect 
the perpendicular GE = k, join EW, and draw EH at right 
angles to EW. 

The velocity of the crank-pin is no longer represented by CL, 
as in Figs. 44 and 45, but by CV. It is found by construction, as 
follows : On crank CL as a diameter construct a circle LUC\ 
from end of acceleration draw a perpendicular OGU to CL\ 
then will CU = CV be the intensity of the velocity of the 
crank-pin C. To find the acceleration Ow of slide, we proceed 
as in Figs, 44 and 45, except that a circle is described with CB' 
as a radius instead of CB^ 

In Fig. 46 the distance OC 3.nd Ow, respectively represent the 

* There is still another way of finding point S (and consequently point iv) 
that is always available. 

Draw VB^ perpendicular to stroke IVIVsLud prolong it till it cuts the direc- 
tion H^C of the rod at ^'. Join JV and F, make ^^T parallel to ^Fand 
draw TS perpendicular to IVN". 

For the similar triangles B^CV and ^CZ'give 

CS:CT: : CB' : CV or CB' X CT = CS X CV, 
and the similar triangles ^^CT'and IVCV give 

CB' XCV = CTX CIV. 

Multiplying these two equations and canceling, we get, 



CB' 
^^ ^ cW 

and this is similar to the result found for BF', Fig. 16, pp. 41-43. 



I02 RESULTANT OF WEIGHT OF ROD AND ITS FORCE OF INERTIA. 

accelerations of crank-pin C and wrist-pin W. The line Cw is 
the reduced image of the center line CW of the rod. We may 
consider this rod to be an exaggerated specimen of the Westing- 
house type in that its center of gravity G is off the center line 
CW oi the rod. We may regard the triangle CHGW as the 
representation of the rod and proceed to construct a reduced 
image of it on Cw as a base. To do this lay off Cw = Cw, 
draw w'U parallel to WH and connect C with G. The triangle 
Ck'^'w' is evidently similar to CHGW. Revolving it back to 
Cw as a base we get triangle Chgw as the desired image of the 
rod, and Oh, (9^ will be, in direction and intensity, the accelera- 
tions of points H, G, respectively. Through //draw ///parallel 
to Oh ; it will cut at / the direction WI of the acceleration of the 
other mass at W. Through / draw, parallel to Og, the line /AT; 
it will be the actual location of the resultant force of inertia of 
the rod, and its intensity, as before, is equal to the product of the 
whole mass of rod X acceleration g oi the center of gravity. 



Combination of Weight of the Rod with its Force of 

Inertia. 

An exact determination of the pressure exerted by rod against 
the crank- and wrist-pins must include Its weight. 

We will, therefore, find the resultant of the weight and of the 
resistance due to inertia. As the inertia is measured by mass 
times acceleration, the expression of the force of inertia becomes : 

Inertia of rod = — X ^ X %= F, (71) 

g OC R 

W z= weight of rod in pounds ; 
V = velocity of crank-pin in feet per second ; 
R = radius of crank in feet ; and 
g = acceleration due to gravity := 32.16 feet. 
^ g* OC = accelerations of center of gravity and of crank-pin 
as given by Figs. 44 and 45. In the case of Fig. 46 we must 



RESULTANT OF WEIGHT OF ROD AND ITS FORCE OF INERTIA. IO3 

substitute in the above expression for OC the value O^C. As 
the forces connected with engine are usually reduced to pressure 
per square inch of piston, the total inertia above given and the 
weight I^F should be divided by the area A. 




Fig. 47- 

In Fig, 47 pass a vertical GQ through the center of gravity G 
till it cuts the direction IK of the force of inertia of the rod. 
Make QS = weight of rod per square inch of piston, and QK 
= inertia-resistance of rod per square inch of piston. Complete 
the parallelogram ; we get the diagonal PQR as the desired re- 
sultant of weight and inertia. This resultant must be overcome 
by the combined action of the pressures of crank- and wrist-pin. 

Example: Harris- Corliss engine, 26^ X 60. R. p. m. = 60. 
Length of rod = 150 inches. Weight of rod = 1,200 pounds. 
Weight of other reciprocating parts ^ 1,300 pounds. Radius 
of crank = 2.5 feet. v = 1571 f^^t per second, v^ = 246.74. 
Principal polar radius of gyration = k = 4S inches. Distance 
of center of gravity from wrist-pin = 78 inches. Distance 
of center of oscillation from center of gravity = 29.57. As- 
suming crank at 60° and drawing the mechanism to a scale of 
an inch to the foot, we get Og- z= 1.595 inches and OC = 2.5 
Substituting in the above formula and dividing by A we 



inches 
have 



A 



] = 4.38 = QK and 



"a 



= 2.24 = QS. 



I04 PRESSURES ON FRICTIONLESS CRANK- AND WRIST-PINS. 

Exact computation* gives 

F 

A = 4-37. 
which differs but 00 1 pounds per square inch from the result 
obtained by the graphical method. Considering the small scale, 
one-twelfth, employed, this shows the practical excellence of the 
graphical method. 

In the above example the weight is one half the inertia-re- 
sistance, but that is because the engine is large and of compara- 
tively slow speed. In small, high-speed engines, the proportion 
is much less. Thus, in a A^. K ^. S.P. engine, lox 12, with rod 
of seventy pounds and 300 r. p. m. [i. e., same crank-pin speed 
as in the large engine) we found 

W" F 

— - =0.89 and ^ = 9.21 

or weight = one-tenth inertia. In the latter case the resultant 
will differ but little from the inertia in direction and intensity. 

F. 

Pressures at Crank- and Wrist-pin when there is no 
Friction at the Pins. 

When there is no friction at the pins, let Pc and P^ represent 
respectively the crank-pin and wrist-pin pressure. Let line Iq, 
Fig. 48, represent the direction of the resultant of the weight and 
inertia of rod, and let the intensity of this be assumed equal to 
F^=i Ip = kh. It is evident Pc and P^ combined must equal this 
resultant, and if one of these forces (say P^) is known, the other 
Pc can be found by the parallelogram of forces. To obtain P^ we 
must satisfy two conditions, one of which is that P^ must do its 

* The computation was made by the help of formulas developed by Profs. 
Jacobus and Webb, in a paper published in Trans. Am. Soc. Mech'l. Eng., 
vol. xi. The data for the above example was obtained from the same source 
(see p. 498). 



PRESSURES ON FRICTIONLESS CRANK- AND WRIST-PINS. 



105 



part in producing the resultant F^ acting along Iq, and the other 
that I\ must balance the resultant of the guide reaction 6^ and the 
driving force K of the piston. This force K is the difference 
between the effective steam-pressure P and the inertia F of pis- 
ton, piston-rod and cross-head. 




Fig. 48. 

Through W draw Wz equal and parallel to F^ = Ip = kh. 
Join any point / on Iq with C and W. Draw zv and Wv, respec- 
tively parallel to /(T and IW. Through the intersection v draw 
vS parallel to center line CW of rod, which line also represents 
direction of internal stress, as Cand IV are the points of applica- 
tion of the forces. We will show later that this parallel tS is the 
locus of the ends of all forces applied at IV that satisfy the first 
condition mentioned above. Now lay off Wd = A" = P — F^ 
= driving force and draw <^/ parallel to guide reaction G. The 
line /^/ is the locus of the ends of all forces applied at W that 
satisfy the second condition. The intersection t of these two loci 



I06 PRESSURES ON FRICTIONLESS CRANK- AND WRIST-PINS. 

St and dt, gives Wt = P^, the force desired. Prolong Wt to 
point ^ on line of resultant, then making M = F^ and kf :z=zWt 
=^P^ and completing the parallelogram, we get the crank-pin 
pressure Pc =^ kg and the rotative effect z= P^ X OL. 

For convenience of expression we will speak of the two com- 
ponents of each pin pressure as internal-stress component and 
inertia-component. The latter expression is not strictly correct as 
the second component not only overcomes the inertia but also sustains 
the weight of the rod, but this will not at all affect the accuracy 
of our results. 

Since the force Pc passes through crank-pin center C and may 
have its point of application anywhere on its line of direction 
without affecting its accelerating capacity, we will assume this 
point to be at C, and since force P^ passes through wrist-pin 
center W, we will, for the same reason, take its point of applica- 
tion at W. This will make the center line CW of the rod the 
direction of the internal stress caused by either P^ or P^. 
Although the intensity of this internal stress varies as the inertia 
components Pc and P^, are varied, nevertheless there is but one 
pair of forces Pc, P^ that will satisfy the conditions of this prob- 
lem. To prove this it is only necessary to show that there is 
only one force P^ that will satisfy the conditions at the wrist-pin. 
One condition is that it shall be in equilibrium with the known 
force K = P — Fj^ = Wb and the partly known reaction G of the 
guides. By drawing through b a parallel bt to the known direc- 
tion of G, we get one locus for the end of the force P^^j. Another 
condition that P^ must satisfy is, that it shall be the resultant of 
the internal stress acting along C W and the corresponding 
inertia component acting in any arbitrarily chosen direction. Let 
Wl be any such direction, then will CI be the direction of the 
corresponding inertia component that acts at the crank-pin C. 
If kh =^ Ip = Wz = F^ is the known resultant of the weight and 
inertia of the rod, then Id = Wv is the intensity of the inertia 
component acting along the assumed direction Wl, and le = vz 



PRESSURES ON PINS WHEN THERE IS FRICTION. TO/ 

is the inertia component along CL Now draw through v a 
parallel Svt to the inner stress CW; it will divide the triangle 
VVvz into two triangles WSv and zvS, respectively similar to 
triangles Wn/ and hiC. Hence 

liW X SW = 2^ X In = nC X Sz 
and 

-— =: —-- = constant (72) 

for this position and motion of the rod. Since Ws is also con- 
stant for this position of rod, the parallel Svt will pass through 
the same point 5 whatever the direction and intensity Wv of the 
inertia component. Therefore, the intersection of Svt with the 
locus bt will give Wt as the only possible value of P^^ for this 
position and motion of the rod. 

Prolonging P.^^ to k and drawing kC and also tz, we get P^ 
1= kg =/^. It is also evident from the figure that the reaction 
G -.=: bt oi the guide is independent of the direction and intensity 
of the inertia-component. 

G. 

Pressures at Crank- and Wrist-pins when there is 
Friction at the Pins. 

When there is friction at the pins let Pc/ and P.^„f represent, 
respectively, the crank-pin and wrist-pin pressures, and let GKL, 
Fig. 49, represent as before the location of the resultant F^ of 
weight and inertia of rod. We will again suppose F^ = ON 
= KQ = PO = TL to be known. In this, as in the former 
case, P^ and /^^/ combined must be able to sustain the weight 
and overcome the inertia of the rod. 

Moreover, Pc/ must be tangent to a circle described from C as 
a center with ^r^^ as a radius, and P«^ must be tangent to a 
similar circle about W with wx.„ as a radius when ^ is coefficient 
of journal friction, and r^, x.^ the radii, respectively, of crank-pin 



io8 



PRESSURES ON PINS WHEN THERE IS FRICTION. 



and wrist-pin. These circles are called friction circles and each 
force is drawn on that side of the friction circle, which will give 
a component opposed to the motion of the eye or pin according 
as the action of the pin or action of the eye is the force under 
consideration. 




Fig. 49. 

In Fig. 49 the friction circles have been greatly exaggerated, 
and the rod given unusual proportions for the purpose of better 
illustration. 

With this modification the force P^^^ must satisfy the same two 
conditions as in the preceding case. But in that case it was 



PRESSURES ON PINS WHERE THERE IS FRICTION. ICQ 

known that P^ passed through the center W of wrist-pin, and 
that point could, therefore, be assumed as its point of application. 
In the present case the possible values of P^jr do not pass 
through such a common point, and we must, therefore, make use 
of trial methods to find /%/-. 

We will first give a method which is, theoretically, the most 
exact, and then another, which is simpler, and, at the same time, 
accurate beyond the needs of practice. 

From each of the series of points G, P, T, on the line GL 
draw a tangent to each of the two friction circles mitg and kip. 
Make distances GN = PQ = TL equal to the known force F^ 
and resolve each of these distances into components along the 
respective tangents. Then with W^as a pole draw Wa equal and 
parallel to GA^ Wb equal to PB^ also Wd equal and parallel to 
7Z>, and so on, thus getting a polar curve eabd, whose vectors 
satisfy the condition that the pin forces shall be capable of balanc- 
ing the resultant F^ The second condition is the same in this as 
in the preceding case, namely, that the wrist-pin pressure P^y shall 
be in equilibrium with the driving force K = WV =^ P — Fj. 
and the given reaction G (known in direction only). Now laying 
off WV = K, drawing Vb parallel to the guide reaction G', and 
joining its intersection b (with the polar curve eabd) to the pole 
W, we get in Wb the exact intensity and direction of the wrist- 
pin pressure P^y, but not its location. The latter is found by 
drawing HPB parallel to Wb and tangent to the friction circle 
p/k. Finally, drawing PBI tangent to friction circle mng of pin 
C, and constructing the parallelogram of forces, we have PB 
= Wb = P^f and PB = Pc/, the exact pin forces desired. The 
rotative force is measured by Pcjr X vw. 

In the simpler and only theoretically less exact method, the 
principal steps are hke those taken in finding P^ and P^, in Fig. 
48. The direction of the internal stress is, however, no longer 
CW, and the determination of a convenient direction for it, con- 
stitutes the main portion of the problem. 
S 



I lO PRESSURES ON PINS WHEN THERE IS FRICTION. 

To illustrate the method, suppose the problem solved and the 
forces Pcf and i^^yjust found to be each resolved into two com- 
ponents, one parallel to GL and passing through C and W, re- 
spectively, and the other in the direction of the internal stress. 
The two components parallel to GL must together just equal F^ 
and, therefore, be capable of sustaining the weight and overcom- 
ing the inertia of the rod. The direction of each of the two 
internal stress components will evidently be HI, and according to 
d'Alembert's principle they will just balance each other. Inspec- 
tion of the figure will show that for friction- circle tangents differ- 
ing but slightly in direction from Pcf and P^y, the direction of 
the internal stress will also differ but slightly from HI. This is 
the case even in the present figure, where the rod is very short and 
the friction circles excessively large. Now draw P'^^ parallel to 
P^f, and tangent to the friction circle plk, the location of this tan- 
gent will differ but little from the actual force P.^/. For the same 
reason F c parallel to Pc and tangent to qm7i, will differ but little 
from Pcf. Joining the intersections /^and E (of PJ and P^' with 
the parallel components WM 3ind Ct, respectively), we get in FE 
a line whose direction is nearly parallel to the exact direction of 
the internal stress HI 

Our procedure in finding /^^^/is, therefore, as follows : We get 
the component WS as it was got in Fig, 48 : we next draw through 
5 a parallel Sb to the line FE, just described ; then will Sb be a 
locus of the end b of the desired force Wb. But another locus 
is needed to completely determine b ; this is given by the con- 
dition that Wb = P.^f shall be the resultant of the driving force 
K = P — F^ and guide reaction G\ Therefore, laying off WV 
= Kand drawing Vb parallel to G' we get a second locus whose 
intersection b with the first locus Sb gives the desired intensity 
Wb of the force P,^/, for Wb is common to each of the force tri- 
angles WSb and WVb, thus satisfying both conditions imposed 
upon P^f. The exact location of P.^f may now be found by 
drawing /^P parallel to Wb and tangent to the friction circle//^. 
The accuracy of the work may be checked by completing the 
parallelogram OBPB' and seeing if PB=^ Wb. 




f the lines representing 

ounterweight. 

al to X the mass of the 

lass of the reciprocating 
nass of the reciprocating 
X the mass of the recip- 
es 'e mass of the recipro- 
\ 300- 



and reciprocating parts 
of piston ; hence cen- 
quivalent to ^ of the 

t, are given in the last 

t d, d, d, d, d, etc., see 



foF Resultant of 
IriNG Parts. 



j[ the 
2bntal 
nents 
tia of 
)ving 
ts. 



Sum of the 
Vertical Com- 
ponents of In- 
ertia of Con 
necting Rod. 
Lbs. per W. 



1-33 
2.62 

3.83 
4-92 
5.86 
6.63 
7.19 
7.53 
7.65 

7-53 
7.19 
6.63 
5.86 
4.92 

3-83 
2.62 

1-33 
0.00 



FORCES TENDING TO SHAKE THE ENGINE BED. I I I 

The approximation just given can be still further simplified for 
all cases that may arrise in practice. In such cases the friction 
circles are very small and far apart, and the tangent RJ to both 
friction circles is scarcely distinguishable from the direction HI 
or EF. We may therefore treat this tangent RJ 3.s the direction 
of the internal stress, draw Sd parallel and then proceed as above 
in the determination of Wd and location of its equal P^y. The 
inaccuracy involved in this last, approximate, construction is much 
less than that connected with the best determined coefficients of 
friction. 



H. 



Determination of the Forces Tending to Shake the 
Engine Bed. 

The constant forces acting on the engine bed, like the weight 
of fly-wheel and pull of belt, tend only to shift the bed. It is the 
variable forces that tend to shake it, namely, the effective steam 
pressure P against cylinder covers, the pressure of the cross- 
head against the guides and the pressure of the crank-shaft 
against its bearings. By constructing a polygon of forces, it can 
be shown that the force tending to shake the engine bed is the 
resultant of all the unbalanced accelerating forces acting on the 
different links of the machine. In other words, the shaking force 
is the resultant of the forces of inertia of the moving parts. 

The same conclusion is reached from general considerations. 
Suppose the forces of inertia to be replaced by their resultants 
acting as external forces. Then the whole system of external 
forces will be in equilibrium, the moving pieces may be considered 
at rest for the instant, the whole machine acting as one rigid 
piece. As the steam then acts equally in opposite directions 
upon the machine, its shifting and shaking influence are both nil. 
The internal forces at the bearings occur in pairs of equal and 




a, a. a, a, curve drawn through the extremities of the lines representing 
the shaking forces, assuming the engine to have no counterweight. 

i, b, b, b, curve if counterweight employed is equal to ^ the mass of the 
reciprocating parts. 

c, c, c, c, curve for counter weight equal to '/i the mass of the reciprocating 

d, d. d. d, curve for counterweight equal to >i the mass of the reciprocating 



A, B, C. D. 



;forc 



A-eight equ 



: then 



3 of the r 



;for 



the entire mass of the 






eating parts. 

Bore, lo". Strolte, 12". Revolutions per minute, 300. 

Cylinder at right of diagram. 

Other dimensions given under Case I, Table XI. 

Centrifugal force of total weight (160 lbs.) of rod and reciprocating parts 
when connected at crank-pin is 31^165, per C of piston: hence cen- 
trifugal force of counterweight, when its mass is equivalent to )i of the 
total weight (160 lbs.) is >i X S'-^S = "gSS lbs. per Q"- 

For this engine the coordinates of curve a. a. a. a. are given in the last 
two columns of the following table. 

For method of constructing other curves, b, b, b, b, d. d, d, d, etc.. see 
p. 113. 

Horizontal and Vertical Components of Resultant of 
Forces of Inertia of Reciprocating Parts. 







Horizontal 


Sum of the 


Sum of the 




Acceleration 


Component of 
Inertia of Con- 


Horizontal 


Vertical Com- 


Angles. 


of the Recip- 


Components 


ponents of In- 




necting Rod. 
Lbs. per Q". 


of Inertia of 


ertia of Con 






all Moving 


necting Rod. 










Lbs. per D". 


0° 


20.50 


14.67 


35-'7 


0.00 




20.06 


14.40 


34.46 


"■33 




18,76 


13.61 


32.37 




3°: 


16.69 


12.34 


29.02 


3.83 




■3-99 


.0.64 


24.63 


4.92 


50° 












7-33 


6-33 


1366 


6.63 




3.74 


3.90 


7.64 


7.19 


K 




1.42 


1.70 


7.53 




-2.97 




—3-99 


7.65 


100° 


-5.85 


—3-33 


-9.<8 


7-53 


no" 




—5-44 


-1372 


7.19 


120° 


—10.24 


—7.33 


-17.57 




,30° 


— 11.79 


--^11 


—20.74 




140° 


— ■2-93 


—23.21 


4.92 


150- 


— 13-74 


— 11.32 


-25.06 


3.83 






-12.07 


-26.34 




170° 


-'4-55 


-12,51 


-27.06 


1.33 




— 14.64 


-12.65 


—27.29 





112 DIAGRAM OF FORCES TENDING TO SHAKE ENGINE BED. 

Opposite forces, and their shaking influence is therefore likewise 
nil. The only external forces remaining are the resultants due 
to inertia, the action of gravity and the pull of the belt. The 
weights are constant forces, but their points of application change 
with the motion. In stationary engines we may neglect the 
shaking influence due to this cause. As regards the pull of the 
belt Mr. W. Willis has shown by his experiments that the sum 
of the two tensions is not a constant quantity, as is generally 
assumed, but increases with the load. If we suppose the load 
constant the belt pull will also be constant, and will only tend to 
shift the bed without shaking it. The only external forces re- 
maining are the variable forces of inertia, and these do tend to 
shake the engine frame. In the case of the engine, therefore, 
the shaking forces are, the resistances due to the inertia of the 
purely reciprocating pieces (piston, piston-rod and cross-head), 
to the inertia of the connecting rod and to the inertia of the crank 
disc. The first and last are easily obtained by computation, and 
the second can be obtained by the construction shown in Fig. 44 
or Fig. 45. 

I. 

Diagram of Forces Tending to Shake Engine Bed. 

Fig. 50 is such a diagram prepared by Prof D. S. Jacobus and 
published in the Trans. Amer. Soc. Mech'l Eng'rs, Vol. XI. It 
contains a series of polar curves, aaa, bbb, ccc, etc., whose 
vectors, drawn from the pole or center F, represent in direction 
and intensity the resulant forces of inertia of all moving pieces. 
The data for the construction of the diagram are given on the 
same page, and under Case I, Table XL The table accompanying 
Fig. 50 contains the components or the force of inertia for this 
particular engine, and is in convenient shape for use. The last 
column contains the ordinates, like C^C^, of the polar curve aaaa, 
and the last column but one the abscissas, like VC^, of this same 
polar curve. As this polar curve aaaa, corresponds to the case 
of no counterweight, its vectors represent in direction and inten 



DIAGRAM OF FORCES TENDING TO SHAKE ENGINE BED. II3 

sity the resultant of the forces of all moving pieces, except the 
crank. The remaining curves bbbb, cccc, dddd, etc., can there- 
fore be directly deduced from curve aaaa, by laying off from 
each point a, and parallel to the crank position corresponding to 
this point a the centrifugal force of counterweight on crank disc. 

For instance, when crank is at 40°, the vector VC^ will be the 
resultant of the inertia-forces of all the reciprocating parts. To 
get a point on curve bbbb, for the same crank position of 40°, we 
lay off CJ) rr: 34^ X 31.25 = 7.81 parallel, but opposite to VC . 
To get a point on curve dddd, for this crank angle of 40°, we lay 
off on the same line CJ)d =^ ^ X 31.25 = 19.53 lbs. per □'', 
and so on for the other curves. 

As the vertical components of curve bbbb are very small, its 
small amount of counterweight would be suitable for cases in 
which the shake occurs more readily in a vertical than in a hori- 
zontal direction, as when a horizontal engine is placed on an 
upper floor of a building. On the other hand the horizontal 
components of curve eeee, are small and the corresponding, 
heavier, counterweight is preferable when the horizontal forces 
produce most shake, as in engines set on tall foundations. 

The diagram just described is suitable for stationary engines, 
and can readily be constructed for such cases as soon as the ac- 
celeration of the center of gravity of rod has been found graph- 
ically, (see page loi, Eq. 71 and Fig. 44 or 45). We have then 
to multiply this acceleration by the mass of the rod per □'' of 
piston, and resolve this force into two components, one horizon- 
tal and the other vertical. Of these two components add that 
which is parallel to stroke to the accelerating force (per □'' of 
piston) of the piston, piston-rod and cross-head. This sum will 
be one of the coordinates of the curve aaaa (Fig. 50). The re- 
maining component of the rod will be the other coordinate of 
curve aaaa. The influence of counterweight can then be added 
in the manner already described in connection with Fig. 50. The 
diagrarft represents the shaking forces acting on engine bed in 
the vertical and axial of the engine. 



114 DIAGRAMS OF PRESSURES AT WRIST- AND CRANK-PIN. 



When the engine frame is suspended on 
springs, as in locomotives, the exact location 
of the total accelerating force of rod should 
be taken into account. The exact location 
of the weights is also of consequence in de- 
termining the oscillations of the locomotive. 
Indeed, in this case it would be well to com- 
bine separately the force of inertia and weight 
of each moving piece and then combine these 
resultants into a total resultant with its point 
of application on the line of stroke and lay 
off each of these resultants from its own point 
of application and with its own direction and 
intensity. 

Construct the curve aaaa for the engine 
to be designed, determining the inertia of 
the connecting rod by the constructions 
given in Fig. 44 or 45. For method of 
ascertaining the radius of gyration experi- 
mentally or by computation, see Appendix. 



J- 

Diagrams of Pressures at Crank- and 
Wrist-Pin. 

Rapid changes in the directions of the 
forces acting at the pins are approximately 
indicated by such diagrams as Fig. 27. They 
do not necessarily indicate reversals, this 
term being understood to mean an instanta- 
neous change of pressure from one side to 
the other of a pin. If the pin had been worn 
into an oval shape then it is possible that 
the changes shown in Fig. 27 might be ac- 
companied by shocks. 



DIAGRAMS OF PRESSURES AT CRANK- AND WRIST-PIN. 



15 



In Fig. 51, the pressure of connecting-rod on wrist-pin is 
shown in direction and intensity for the engine whose data are 
given under Case I, Table XI, when friction, weight and inertia 
are all taken into account. These forces are laid off from what 
is equivalent to a time base, namely, the developed crank-pin 
circle. Rapid changes of direction of pressure occur between 
140° and 150° for both forward and return strokes. It is not 
however instantaneous, continuous contact being preserved be- 
tween the pin and its bearing. 

But between slide and its guides the contact very suddenly 
changes from one side to the other in the neighborhood of 150° 
forward stroke and 160° return stroke, provided the upward, 




— scale:.— 

O 20 40 

lim i L-J-JLj-Jl-I-J 

Fig. 52. 



Il6 DIAMETER AND WIDTH OF BELT PULLEY. 

vertical, component or pressure of wrist-pin is sufificient to over- 
come the weight of cross-head and that part of piston-rod which is 
sustained by the sHde. In such a case a pressure of shde against 
guide would suddenly change from a downward to an upward 
pressure and be accompanied by a more or less violent shock. 

In Fig. 52 the pressure of connecting-rod on crank -pin is 
given in direction and intensity for the same engine and condi- 
tions assumed in Fig. 51. Between 140° and 160° of forward 
stroke and 150° and 170° of return stroke a rapid change of 
direction of force occurs, but continuous contact between pin 
and eye is maintained even then and no shock will occur at 
crank-pin under these particular circumstances. At the inner, or 
front dead point A there is a sudden change of pressure on 
crank-pin, which may cause tremor in the rod but no shock. 

IX. 
Diameter and Width of Belt Pulley. 

Generally in high speed engines the fly-wheel acts also as belt 
pulley. In such a case the diameter is determined by the data 
given on pp. 75 and 76. When the diameter is not fixed in this 
way or by some special condition and there is thus left some 
choice, it should be so made as to give the belt a high speed, 
even a mile a minute is permissible, for the belt becomes nar- 
rower and the efficiency increases with the speed, the percentage 
of slip and the journal friction being smaller at the higher speed.* 

*See pp. 347-360, also pp. 568 of Vol. VII. Trans. Amer. Soc, M. E. Mr. 
W. Lewis in experiments made for Sellers & Co., found that with vertical 
belts, the sum of tensions while running might increase to | of the sum 
existing when belt is stationary; that with horizontal belts the sum of 
the tensions might increase up to the breaking limit. In both cases this 
increment in the sum is accompanied by increase of slip. In one extreme 
case, with the permissible slip of 3 feet per minute, the increment of the 
sum was as high as 100%. This does not call for any modification of the 
results given above however, as the coefficient of friction is dependent upon 
the slip and the assumption of a 0.27 coefficient means a definite slip and a 
ratio of tensions, in tight and slack sides, of about 2.4. 



DIAMETBR AND WIDTH OF BELT PULLEY. 11/ 

When belt speed is known, its width can be ascertained from 
the following table, provided the belt is laced and connects two 
equal pulleys, 

Belt speed = looo, 2000, 3000, 4000, 5000, in ft. per niin. 

Horse -Powers ) ) per n" of 

. , r= 50 9.8 14.3 18.3 20.8 y^ , , 

transmitted j ^ ^ ^0 | ^^^^ 

This table assumes a permissible working stress of 300 lbs. to 
the square inch at laced joint and a coefficient of friction = 0.27 
for leather on cast iron. For a riveted joint the working stress 
would be larger say from 400 to 550, and the H, P. trasnmitted 
per Q'' of belt from L4 to 2 times greater. Some authorities 
use a coefficient of friction of 0.40, which would make the H. P. 
per □'' in the above about ^ larger, but it would be accompanied 
by a velocity of sliding of belt on pulley of much more than 
three feet per minute. 

When pulleys connected by belt are unequal, account must be 
taken of the arc of contact on the smaller pulley, by multiplying 
the horse-power given above by the fractions in this next table: 

Arc of contact = 120° 130° 140° 150° 160° 170° 180° degrees. 
Multiplier = 0.79 0.83 O.87 O.91 0.94 O.97 I. GO 

The maximum horse-power that the engine is likely to develop 
must then be divided by the H. P. per □'' of the belt found by 
these tables and the result will be the area cf the cross-section 
of the belt. The area is given because there is considerable 
variation in belt thickness, and the mere designations of single 
and double do not necessarily mean i^" and tV' thickness re- 
spectively. Of course to get the belt width the area of belt is 
divided by the thickness. 

In this engine assume that a double belt is to be used on the 
fly-wheel as a belt pulley and that the arc of contact on the 
driven pulley is 165°. 

Many high-speed engines, like the center-crank type, are pro- 
vided with two fly-wheels or belt pulleys and when both are used 



Il8 DETERMINATION OF DIAMETER OF CRANK SHAFT. 

for the purposes of transmission there will be a corresponding 
reduction in the area or width of each belt. 

It is usually in connection with Fly or Band Wheels that we 
must distinguish between right and left-hand engines, between 
those that " run over " and those that " run under," Standing at 

the cylinder end and looking towards the crank, a< i a. u ^i f 

will have the fly-wheel and main bearing to the-^ ^r. > of the 
axis of the cylinder, and still retaining the position at the cylin- 
der, if the top of the fly-wheel moves ] t H [the observer, 

,, . . • J ^ ( run over ) 

the enmne is said to < j ^ . 

^ ( run under J 



X. 

Graphical Determination of Diameter of Crank Shaft. 

The problem is one of compound stress, and has been treated 
by Reuleaux in his Konstrukteur. Both bending and twisting 
forces strain the shaft. The bending forces are: weight of the 
fly-wheel, pull of the belt, the centrifugal force of the counter- 
weight on crank disc, the pressure of connecting-rod against 
crank-pin and the reaction of the bearings. The twisting force 
is the moment of connecting-rod against crank-pin. 

As the bending forces are not parallel, and do not all act in 
the same plane, the first step is to combine these separate bend- 
ing moments into one resultant bending moment. The next 
step is to combine this resultant bending moment with the twist- 
ing moment so as to form either an equivalent bending, or an 
equivalent twisting moment. In the present case we will reduce 
all the moments to one equivalent bending moment. 

The first step may be simplified by certain preliminary reduc- 
tions and approximations. The weight of fly-wheel and the pull 
of the belt act in the same plane, and can be reduced to a single 



DETERMINATION OF DIAMETER OF CRANK SHAFT. II9 

force. The belt pull is the sum of the tensions on the tight and 
slack side of the belt, and this sum is not constant, but in- 
creases, while running, with the load. But this does not call for 
special calculation at this place, because this increase in the sums 
of the tensions is limited by the slip, and this was assumed at 3 
feet per min. as a maximum. We will therefore assume the 
sum of tensions to be i ^ times the tension on the tight side 
of the belt when transmitting the maximum power. This, under 
our assumptions (p. 17) will generally be somewhat in excess of 
the true sum. Assuming that shop shaft and cylinder are on 
opposite sides of crank shaft, and that the belting makes an 
angle of 30° with the horizon, we can now combine the belt pull 
with the weight of the fly-wheel by the parallelogram of forces 
and note the angle made by the diagonal with the horizontal. 
The centrifugal force of the counterweight and the pressure of 
the connecting rod against crank-pin do not act in the same plane 
and should be given separate diagrams, if great exactness is de- 
sired. But such a refinement is here entirely unnecessary, and 
we may approximate by assuming them both to act in some 
convenient, intermediate, plane which is nearest to the larger of 
these two forces. These two may then be combined by the par- 
allelogram of forces, again noting the angle the diagonal makes 
with the horizontal. 

The bending forces have thus been reduced to two forces and to 
the reactions of the bearings. We now construct by the methods 
of graphical statics a diagram of bending moments for each of 
these forces. 

Suppose that in Fig. 53 the scale of distances is %, or 3" to 
the foot, and the scale of forces is, say 200 lbs. to the inch. The 
polygon of forces due to the fly-wheel resultant found above, is 
ABC, the length AB representing the reaction of the out-board 
bearing and BCth.^ reaction in the main bearing, the two reac- 
tions having the ratio MN -^ NQ. Choosing O in the polygon 
of forces on the horizontal through B, the closing line AD of 
the equilibrium polygon ADGF\w\\\ also be horizontal, and thus 



I20 DETERMINATION OF DIAMETER OF CRANK SHAFT. 

convenient for subsequent combinations. The distance OB of 
the pole from AC is chosen so as to give a convenient factor 
for multiplying the vertical chords or intercepts HI of diagram 
of moments ADGF. We know from the principles of graphical 
statics that the product of the intercept HI by the constant OB 
measures that part of the bending moment at cross section KL 
of shaft which is due to the forces on the fly-wheel. As OB is 
constant, the intercepts themselves may be taken to represent the 
bending moments of the cross-sections of the shaft directly above 
them. The numerical value of any bending moment is found by 
multiplying the proper intercept by the distance OB X 200 X 4, 
the last two factors respectively representing the scale of forces 
and the scale of distances on drawing. 

If we take OB equal 4 inches, the product of these three factors 
= 3200, which is a convenient factor with which to multiply 
directly the intercepts ///obtained from ADGF, the diagram of 
moments. 

In like manner for the second resultant, due to counterweight 
and crank-pin pressures, we find the polygon offerees A'B' CO' 
and the equiHbrium polygon D' E O . 

This second resultant is constantly varying in direction and 
intensity with the rotation of the crank. It is greatest at the 
beginning of the forward stroke. But this does not correspond 
with the piston position at which the maximum twisting moment 
exists. As the greatest stress on the shaft probably occurs near 
this position, we shall suppose the diagram of moments due to this 
second resultant to be found for the crank position corresponding 
to the maximum twisting moment. The vertical intercepts of 
polygon D' E C represent the bending moments, which are due 
to the action of this second resultant and its numerical value for 
any cross-section is found by multiplying 3200 the intercept of 
this diagram immediately below the cross-section in question. 

These two diagrams of bending moments are both placed, for 
convenient combination, in the plane of the paper, though the 
forces calling them forth are neither parallel nor do they act in 




Fig. 53. U 



122 DETERMINATION OF DIAMETER OF CRANK SHAFT. 

the same plane. As the intensities of the bending moments are 
represented by linear quantities and their directions are those of 
the forces producing them, they may be combined like forces, by 
means of the parallelogram of forces. This we will now proceed 
to do, being careful to take into account the angle between the 
forces. It will probably often happen that this angle is a small 
one, in which case the intercepts of the two diagrams may, with 

little error for the present purpose, be directly < , ♦- h i 

when the directions of the bending moments developed at any 

f aHke 1 
cross-section by the two resultants are < ,., >. 

When the angle /'/// = t, included between the arrowheads 
of the forces is large they should be combined on the principle 
of the triangle of forces. In the figure this is illustrated by lay- 
ing off the length N^f of one diagram on ///and joining ^and 
/. We now take this line // of the second diagram and lay 
it off below the horizontal XX making it equal to RS, which 
is located on the vertical SRIHH' passing through the 
assumed cross-section KL. Repeating this process for every 
other cross-section we get a third (the lowest) diagram XRXZSZ 
whose intercepts represent the resultants of the purely bending 
moments attacking each cross-section. ^ 

The next step is to combine this resultant bending moment 
Mb with the twisting moment Mt, so as to get an equivalent or 
ideal bending moment Ma. The formula connecting these three 
quantities is : 

Mh = yiM, -^ s/q VM,^ + M,\ (73) 

Before undertaking the graphical construction of this formula 
we will construct the diagram of twisting moments to the same 
scale as the bending moments. The twisting moment varies 
for the different crank-positions, but we will at once assume its 
maximum value Pc/R. To have the intercepts of the diagram 
of twisting moments on the same scale we must make 
Pcf X R = O' B' X y. Making, in one of the polygon of forces, 



DETERMINATION OF DIAMETER OF CRANK SHAFT. 123 

C'V =^Pc/ and 0' V = R = fg, we get, by similar triangles, the 

distance VW = ^ ^ = y. Now lay o^ la = y above the 

O'B' 

horizontal CD' and through a draw ab horizontal. The in- 
clined line bee can be taken to represent the varying twisting 
moment of crank-shaft within the fly-wheel hub. The vertical 
intercepts of diagram abcC represent the twisting moment on the 
same scale as the other moments. 

We can now return to the graphical construction of the last 
formula. The last term containing the radical can also be written 



The first term Sy^ Mi,, under the radical is found by taking S^ of 
each vertical intercept of the diagram XRXZ USZ. For in- 
stance, Rs ^^ Y^ RS. The broken line in this diagram represents 
this division of RS =^ Mi, into ^ and y% segments. In like 
manner ^ .'lf^=y^'=^/^can be found. Making ^;;2 = y^ Mt=^H'i, 
the hypothenuse ms{=sT) of the right-angled triangle mRs is 
the graphical equivalent of the last, radical, term of our formula. 
Adding it to Ss ^ ^ M^ we get ST = M^. Its numerical 
equivalent, with the scales assumed above, is Mbi = length of 
5r X (9^ X %; X 200 inch lbs. This value of M^i is to be 
substituted in the general formula for shafts subjected to bending 
stress 

d= 2.17 31 .^inches, (74) 

where /^ is the permissible working stress in lbs. per □'' when 
shaft is subjected to bending stress. The result may be checked 
by the following empirical formula representing high-speed en- 
gine practice, for diameter of main, crank-shaft, journal 

d = 0.44 D ^ %; (75) 

here D represents the diameter of the cylinder. 

Having indicated the general method of procedure it will now 
be easy to apply the graphical solution to cases in which the fly- 
wheel overhangs the bed. 



124 



PLANE OF DIVISION OF THE BRASSES. 



XI. 

Determination of Plane of Division of the Brasses and 
Length of the Main Bearing. 

The plane of division of the brasses should be at right angles 
to the resultant of all those pressures which exist when the en- 
gine is running with its average load. These pressures are com- 
ponents of the weight of shaft and fly-wheel, pull of belt, centri- 
fugal force of counterweight and the thrust or pull of connecting- 
rod. For each crank position these four components may be 
combined by the polygon of forces so as to give the resultant 
force on bearing. 




Fig. 54- Fig 55- 

In Figs. 54, 55 and 56 these forces are shown and their force 
polygon. For a given load, speed and engine, RB and BWrnay 
be regarded as constant in direction and intensity, consequently, 
i? PF may have an invariable position on the diagram Fig. 56. 
As the action of the counter-weight is constant in intensity and 
opposite to the crank in direction, we may take W as the center 
of a circle described with Wrsi as a radius. From the points of 
this circumference we draw tjj T' equal and parallel to the (reduced) 



PLANE OF DIVISION OF THE BRASSES. 



125 



values of Pc/ corresponding to the different crank positions, then 
will i? 7" represent the resultant of the two pressures exerted by 
the shaft against its main bearing. Fig. 56 shows two determi- 
nations, RT and RT , of these resultant pressures against the 
bearings for the 30° and 60° crank positions. 




Fig. 56. 



The forces in this figure (56) have been chosen at random and 
do not represent any special case. When a graphical determina- 
tion, Fig. 53, of the diameter of the crank shaft has been made, 
the components of these pressures at the main bearing can be 
found from the force polygons there drawn. A resultant 
should be drawn for each of a series of equidistant crank posi- 
tions and the plane of division of the brasses placed at right 
angles to the resultant of these resultants. Great accuracy is 
not necessary, in most cases inspection will determine best posi- 
tion of plane. 

For the determination of the length / of crank-shaft journal of 
a high-speed engine the following empirical formula, represent- 
ing good practice, may be used 



/= \V,d Ar 



3'^o 



(76) 



126 DETERMINATION OF DIMENSIONS OF STEAM PASSAGES. 

where d is the diameter of same journal determined according to 
another empirical formula, for same class of engines, 

d = 0.44 D -I- yi", 

D representing the diameter of piston. 



XII. 



Determination of the Dimensions of Steam Passages. 

The first dimension to be settled in valve-gear problems is the 
requisite area of the ports, for upon this depends the maintenance 
of the desired steam pressure. The principal factor in its deter- 
mination is the velocity of the steam current and as this varies 
with the piston's motion in different parts of the stroke, we must 
see to it that the minimum cross-section in the passage is adequate 
for every part of the stroke. This part of the problem divides it- 
self into two parts: {a) the ascertainment of the minimum area 
needed at the given piston speed and with given cylinder area, 
and (^) an examination of the cross-sections of the steam passages 
that are really available or effective with the special type used, 
taking into account the contraction caused by the flow past edges 
and the narrowing of ports caused by the motions of the valves 
on each other or on their valve seats. 

This second part presents no theoretical difficulties and we 
shall therefore not examine any special case here, but this should 
be done in the draughting-room when the type of valve has been 
chosen. It is practically important in case there are supplemen- 
tary, or multiple, ports in the main or distribution valve. Allow- 
ance should also be made for the contraction of the steam cur- 
rent when the ports are very narrow, for this effects a very notable 
reduction of the port area geometrically available. We will now 
consider the minimum port area necessary for a given area of 
cylinder and a given piston speed. 



DETERMINATION OF DIMENSIONS OF STEAM PASSAGES. 12/ 

Radinger's criterion of sufificient passage area was whether or 
not the admission Hne of indicator card remained horizontal up 
to cut-off. 

According to Radinger's experiments, 

Area of port bl average speed of piston in feet per sec. , . 

Area of cylinder ~~ A ~ loo 

This rule evidently corresponds to an average velocity of lOO 
feet per second of the current of steam in the port ; but the 
experiments seem to have been made on engines in which the 
maximum cut-off was equal or greater than 0.5, consequently if 
we assume that in the engines experimented upon the ratio of 
connecting rod to crank varied from 4 to 6 we can easily deter- 
mine the maximum permissible velocity of the steam in the ports. 

-r. . connecting-rod ^ ^ , ^, ^- r 

ror when -^ = 4 to we nave the ratio 01 

crank 

Max. velocity of steam in ports ^ ^ 

-7; i — r- TT— ; = 1.62 to \Ac\ respectively. (78) 

Ave. velocity 01 steam m ports ■'^ ^ j \i i 

This gives about 160 feet per second as the maximum permissi- 
ble velocity of the current of steam in the ports when the steam 
passages are long and narrow. Radinger says that when the 
steam passages are short and the cut-off short (for instance, when 
there are separate admission valves for each end of the cylinder) 
somewhat smaller port areas may be employed than would result 
from above rule ; in other words, in such cases the maximum 
permissible velocity of the current of steam may be somewhat 
greater than 160 feet per second.* 

Mr. Charles T. Porter gives the rule that the velocity of the 
current of steam in the short ports should not exceed 200 feet per 

* Some American engineers prefer the rule that the aver age s^&^dixn short 
steam ports should not exceed 150 feet per second, and in the exhaust ports 
should < 125 feet per second. When the steam is admitted and discharged 
through the same passage, some engineers recommend that its cross-section 
be proportioned for the exhaust, preferring an increase of clearance to an 
increase of back pressure. 



128 RATIO OF PISTON VELOCITY TO CRANK-PIN VELOCITY. 

TABLE XIII. 

Velocity of Piston for a Unit of Velocity of Crank-Pin. 

To obtain actual velocity of piston multiply tabular quantity by actual 
velocity of crank-pin. 
Forward stroke is towards, and return stroke, away from, crank shaft. 



Crank Angles. 


Connecting Rod -;- Crank = 


Forw'd. 


Return. 


4.0 


4.5 


5-0 


5-5 


6.0 


00 


5 


175 


.1089 


.1064 


.1045 


.1030 


.1016 


.0832 


lO 


170 


.2164 


.2117 


.2079 


.2047 


.2022 


• 1737 


15 


165 


.3215 


.3145 


.3089 


•3054 


.3005 


.2588 


20 


160 


.4227 


.4136 


.4065 


.4019 


•3957 


.3420 


25 


155 


.5189 


.5081 


•4995 


•4925 


.4866 


.4226 


30 


150 


.6091 


.5968 


•5870 


•5791 


•5724 


.5000 


35 


145 


.6923 


.6788 


.6682 


.6596 


.6523 


•5736 


40 


140 


.7675 


.7533 


•7421 


.7329 


•7253 


.6428 


45 


135 


.8341 


.8195 


.8081 


.7988 


.7910 


.7071 


50 


130 


.8914 


.8771 


•8657 


.8564 


.8488 


.7660 


55 


125 


•9392 


.9253 


.9144 


.9055 


.8982 


.8192 


60 


120 


.9769 


.9641 


•9540 


•9458 


.9390 


.8660 


65 


115 


1.0046 


•9932 


.9842 


.9769 


.9709 


.9063 


70 


no 


1.0224 


1.0127 


1.0052 


.9990 


•9939 


•9397 


75 


105 


1 .0304 


1.0228 


1.0169 


I.OJ2I 


1.0082 


.9659 


80 


100 


1.0289 


1.0237 


1. 01 99 


1. 01 64 


1.0137 


.9848 


85 


95 


1. 01 86 


1.0160 


1.0139 


I.OI27 


1.0109 


.9962 


90 


90 


1. 0000 


1 .0000 


1 .0000 


1 .0000 


1. 0000 


1 .0000 


95 


85 


•9738 


.9764 


•9785 


•9797 


.9816 


.9962 


1 00 


80 


.9407 


.9460 


.9500 


.9532 


•9559 


.9848 


105 


75 


.9016 


•g??' 


.9150 


.9198 


•9237 


.9659 


no 


70 


.8571 


.8667 


•8743 


.8804 


•8855 


.9397 


115 


65 


.8080 


.8194 


.8285 


.8357 


.8418 


.9063 


120 


60 


.7552 


.7680 


.7781 


•7863 


.7931 


.8660 


125 


55 


.6992 


.7130 


.7239 


•7328 


.7401 


.8192 


130 


50 


.6407 


.6550 


.6664 


.6756 


•6833 


.7660 


135 


45 


.5801 


.5946 


.6061 


.6155 


.6232 


.7071 


140 


40 


.5181 


•5Jo^ 


•5435 


•5527 


.5603 


.6428 


145 


35 


•4549 


.4683 


•4790 


.4876 


.4949 


• 5736 


150 


30 


.3909 


.4032 


•4130 


.4209 


.4276 


.5000 


155 


25 


.3264 


.3371 


•3458 


•3528 


.3586 


.4226 


160 


20 


.2614 


.2704 


.2776 


.2821 


.2884 


.3420 


165 


15 


.1962 


.2032 


.2088 


.2123 


.2171 


.2588 


170 


10 


.1309 


.1356 


.1394 


.1426 


• 1451 


.1737 


175 


5 


.0655 


.0679 


.0698 


.0714 


.0727 


.0872 



DETERMINATION OF DIMENSIONS OF STEAM PORTS. 



129 



second. This is what we shall assume as suitable for short pass- 
ages. Representing the actual velocity of the steam or piston by 
Vs, and by ^' the minimum permissible opening of port when the 
crank makes an angle w with the line of centers, we have the fol- 
lowing formula when the steam passages are long and narrow : 

d' X / Vs _.. „ AJ^ 
I 



60 



or^' 



or 



A_ 
7 



= V, : 160 



when the passages are short, we have 

b ' X / _ V^ 
A 200 

A 



oxb' 



A_ 
7 



160 



Vs_ 
200 



(79) 



or 



b' '. 



I 



Vs : 200 



(80) 
(81) 



The opening of the port b' corresponding to the different crank 
angles may now be graphically determined (Fig. 57) as follows : 
Assuming that the area of cylinder A and length of port / are 




£ 

In this diagram OC 



B C 

A _ area of cylinder 
/ ~~ length of port 
Fig. 57. 



130 DETERMINATION OF DIMENSIONS OF STEAM PORTS. 

given, and finding the velocity of the piston from Table XIII 
on page 128, we first draw a series of radii, making the angles 
10°, 20°, 30°, etc., with the line OX. These angles correspond 
to the crank angles. We now describe a circle CD with (9 as a 

center, and — = 06^ as a radius. We must lay off OB on OX 

and equal either to 160 or 200, according as the steam passages 
are long or short. To find the minimum port opening //, cor- 
responding to a particular position OF of crank, we join the 
point B with F (the intersection of the circle CD with the radial 
line OF^ representing the crank angle in question) and then lay 
off OF on OX^ equal to the corresponding velocity Vs. We now 

draw through E a line FG parallel to BF^ then because OF = — , 

OF = Vs, OB z= 160 or 200 

OG : OF=OF: OB. 

A 
OG : -z= Vs : 160 or 200; (82) 

comparing this proportion with the preceding ones we see that 
OG = ^' = minimum opening of port for crank-position OF; 
in like manner all values of d' can be determined and a curve 
OGG'' may be drawn through the extremities of 0G'\ OG, etc., 
which will represent the values of b'; the portion of the radius- 
vector included between the curve OGG" and the pole will 
give the value oi b' for the corresponding crank angle. 

When b' and — - are drawn full size, Vs and 160 should each 

be laid off to same scale, say 20 feet (per second) to the inch. If 

A I 

-— be drawn on an — scale while b' is drawn full size we must lay 

off 160 on a scale of ^ X 20 feet to the inch while Vs is laid off 

A 
* —, it should be noted, is a linear dimension and is expressed in inches. 



DETERMINATION OF DIMENSIONS OF STEAM PASSAGES. I3I 

as before, to a scale of 20 feet to the inch. When main and ex- 
pansion valves are of the ordinary type the value of b' obtained 
by the above diagram may be laid off directly upon the main 
and expansion valve diagrams and compared with the actual 
openings of port given by the latter. But when the valves 
are of the gridiron type having two admission ports intead 
of one, we must halve the values of b' obtained from the 
above diagram and thus reduced lay them off on the valve dia- 
gram for comparison. If the expansion valve is of the piston 

type we must lay off the values of b' on a scale equal to — — 

(d^ = diameter of piston valve), for the minimum opening b" of 
piston-valve port {i. e. the minimum for a particular crank angle) 
must equal 

,'=±-,, (83) 

The full width b'" of port for piston valve may of course be 
found by substituting for b' its maximum value b in the preced- 
ing formula. Formula (83) assumes that the steam passage 
within the main valve has everywhere a cross-section ^ than the 
maximum, annular, port opening effected by the piston valve. 

As regards the steam passage itself (not the port) if it is large 
enough for the exhaust it will be large enough for the admission. 
Its area should therefore be the quotient, 

. Average piston speed (ft. per sec.) ,^ ^ 

100 to 125 

The port opening for exhaust should also be tested, taking 
200 feet per second as the maximum velocity of the current 
that is permissible. 

By plotting the port-openings as ordinates on the piston stroke 
as a base, and connecting them by a curve, the slope of this 
curve where it crosses the base will be a measure of the rapidity 
of cut-off 



132 



DETERMINATION OF DIMENSIONS OF STEAM PORTS. 



Where many valve-gear problems must be solved, as in a col- 
lege draughting-room, two diagrams like Fig. 57 may be drawn 
which will cover the whole range of practice. In both these 
general diagrams the ratio of connecting-rod to crank = 6, but 
the maximum permissible velocity of steam is 2CX) feet per second 
in one diagram and 160 feet per second in the other, correspond- 
ing to short and long steam passages respectively. For these 

A 
general diagrams assume - = 10 and average speed of piston in 

feet per minute {w') equal to 1000. The diagrams will then be 
applicable to any problem, giving the minimum port-openings 

OG (== //) to a scale of i ^ that is OG on the diagram 

^ ^ 10000/ 

vvill be i^??? — the true size. In this way a diagram whose 

w' A 

scale varies with the assumed data can be made to fit all cases. 
To reduce the port-openings given by these diagrams to their 
full-size values on the Zeuner, valve-circle, diagram a reduction 




PORT OPENING GIVEN BY OIAGRAH 
Fig. 58. 



VALVE DIAGRAMS AND DIMENSIONS OF VALVE AND GEAR. I 33 

arrangement, like that shown in Fig. 58, will be found very con- 
venient. The full sizes are taken from horizontal line i.e. 

As regards the thickness of the cylinder walls, Prof W. C. 
Unwin* says, ''{a) that it should be strong enough to resist the 
internal steam pressure ; {b) rigid enough to prevent any sensible 
alteration of form ; [c] it must be thick enough to insure a sound 
casting ; {d) thick enough to permit reboring once or twice when 
worn. Generally other considerations than strength are of so 
much importance, that the empirical rule agrees better with prac- 
tice than a rule making the thickness t depend upon steam 
pressure :" hence when D is diameter of cylinder 

t == 0.02 D -f 0.5 to 0.02 D -f 0.75. {85) 

XIII. 

Valve Diagrams and Dimensions of Valve and Gear. 

We shall assume that the reader is acquainted with the prin- 
cipal functions and common varieties of valves, and also with 
the terms designating the most important dimensions and parts 
of ordinary valve gear. This is usually given in elementary 
text-books on the steam engine. There is such variety in valves 
in the matter of arrangement of steam passages, in the method of 
balancing and in the division or assignment of functions, and all 
this involves so much special dimensioning that it would need a 
separate treatise to do justice to this part of the subject. We 
can here only attempt to point out the principal types and the 
methods of ascertaining the leading dimensions. 

The ultimate object of all valve-gear discussion is to establish 
the relation existing between movement of piston and movement 
of valve. The piston's motion has already been fully discussed 
on pp. 35-39, and a table on pp. 26-27 gives its travel for dif- 
ferent crank positions. After the travel of the valve has been 
ascertained for these same crank positions it will be easy to con- 
struct a diagram giving directly the desired relation. 

* Elements of Machine Design. Eleventh Edition, Part II, p. 33. 



134 KINDS OF VALVE GEARS. 

The valve motion that has been most extensively used in 
practice is that which is known to mathematicians as ''harmonic 
motion," and to engineers as, the motion of the slotted cross- 
head. Almost all valve-motion deviates a little from this, owing 
to the angular motion of the eccentric rod. Of late years a few 
valve-gears have been devised in which the valve's movement 
differs considerably from that corresponding to the " harmonic 
law;" but as yet these gears have not come into extensive use. 
They have usually complicated mechanisms that do not follow 
any simple law in their valve-movements and are therefore not 
readily amenable to scientific treatment. We shall confine our- 
selves almost wholly to the first, widely used, class. Link- 
motions of the common forms do however come within this 
class, but only the Porter-Allen (Fink) form is much used in 
stationary, high-speed, engine work. In this section only this 
latter form of link motion will receive a discussion. The 
Appendix will contain a general, graphical, method for obtaining 
the valve diagrams of the common forms of link-motions. 

The valve-gears subject to the "harmonic law" can be divided 
into two groups : 

a. Gears with but a single valve which slides on a stationary 
seat. 

b. Gears with two valves, of which one slides on the other. 

The first of these will be considered under the two heads of 
invariable steam distribution and variable steam distribution. 

In second group the main part of our problem will be to show 
that the complex mechanism producing the desired relative 
motion of the valve on its seat, is equivalent to the action of one, 
single, eccentric capable of giving the valve in question an iden- 
tical, absolute, motion, on a similar, but stationary, seat. The 
problem before us is therefore mainly a kinematic one. In order 
that we may simplify the usually difficult and complicated parts 
of this subject it will be necessary to review the elementary parts 
and present them in a new light. 



SINGLE- VALVE GEAR. I 35 



Single- Valve Gear. — Invariable Steam Distribution. 

We have already stated that the main object of valve gear dis- 
cussion is to find the relation between piston and valve move- 
ment, and that the piston's position for various crank angles has 
already been ascertained by suitable formulas and tables. As 
the eccentric is nothing but a short crank, a similar relation, of 
course, exists between the valve-slide positions and eccentric 
angles, but with this difference, that in the former case the ratio 
of crank to connecting rod is a comparatively large fraction, say 
1, while in the latter case the ratio of eccentric radius to eccentric 
rod is a very small fraction, often less than ^l. In the former 
case the piston's motion is unsymmetrical in the first and second 
halves of its stroke, while the valve's motion is practically sym- 
metrical on each side of its middle position and for an infinitely 
long eccentric rod would be perfectly so, and this is the assump- 
tion which is always made in this sort of discussions and which 
will therefore underlie all our subsequent work. Moreover, it 
is practically most convenient to estimate piston travel from 
the beginning of its stroke, while valve travel is best esti- 
mated from its center of motion or mid-position, because of the 
symmetry of its motion and of the arrangement of the ports. 
For these two reasons the expressions for piston and valve 
travel are usually different. Just at present we will confine our- 
selves to the construction of a diagram representing the distance 
of the valve or slide from its mid-position at the different crank 
angles and leave the graphical representation of the piston's 
travel to the moment when we shall combine the two movements 
into one diagram and thus exhibit graphically their desired 
relation. 

We shall first treat the slide-valve as a simple, rectangular, 
plate, moving on an unperforated seat and driven by an infinitely 
long eccentric rod. In Fig. 59, let be the center of the engine 
shaft, OC^, OC^, OC^, OC^ different crank positions and OE^, OE^, 
OEy OE^, the corresponding eccentric positions, the eccentric 



136 



SLIDE TRAVEL. 



being set ahead of the crank by a constant angle C^OE^. Let 
NZ be the dead center line of the eccentric OE (which may or 
may not coincide with that of the crank OC), then will OP be 
the mid-position of the eccentric. When the eccentric is in this 
position the valve will also be in its mid-position. When the 
eccentric is in position OE^, the slide will be at the distance 
Er.Sj. from its mid-position and this distance we will call the travel 




Fig. 59- 



of the slide. For the other eccentric positions OE^, OE^, OE^, 
the slide-travel is E^S^, ^3-^3. ^^^^^ respectively. If now we lay 
off this travel ES on the corresponding crank positions while 
the eccentric is to the right of OP and on the prolongations of 
the corresponding crank positions when the eccentric is to the 
left of OP, and making OD, = E,S,, OD, = E,S^, OD^ = E^S^ 
and OD^ = E^S^, we get the curve ODJD^Dfi^, which is called 
the polar diagram of the slide motion. It is evident from the 
method of construction that the chord cut by this curve from the 



VALVE DIAGRAM. 



137 



crank or its prolongation is the travel of the slide from its middle 
position. It is also evident that the slide is respectively to the 

\ ]%t I ^^ ^^^ middle position when the curve cuts the 
{ prolonged } --'^^ ^^ '^^^ '^e slide is going | ^^^^JJ^ } 
its middle position when these chords < , > . 

With an infinitely long eccentric rod, the curve OD^D^D ^D^ 
becomes a circle. To prove this let us take a general case, Fig. 
60, in which dead center line OX of crank makes an angle XOZ 
with the dead center line OZ of the eccentric. From the right- 
hand half of NOZ, and in a direction opposite to the crank's 
rotation, measure the angle ZOE = a between this dead center 
(9Zand the eccentric position OEt\\dX corresponds to any crank 
position OC. Then lay off from the crank OC in the same di- 
rection as the rotation, this angle a = COD. Make OD =z OB 
and on OB as a diameter describe the circle OFD. The right- 
angled triangles OEF and ODF' are evidently equal, hence 
OF' = 0F= ES = valve travel, that is, the circle OFD cuts 
a chord OF' from any crank position (9(7 that is equal to the 




Fig. 60. 



138 RELATION OF VALVE-CIRCLE AND CRANK. 

valve travel ES for that instant. As the polar co-ordinates of 
circle OF^D and of the polar curve ODJD^D^D^ (Fig. 59) are 
alike, the latter is also a circle, and may be called the slide- or 
valve -circle. 

By reversing the steps of this construction we may evidently 
find the position of the eccentric from the crank position, when 
the valve circle is given. Our rule then is, to measure from the 
crank, in the direction of its rotation, the angle a included be- 
tween it and the diameter of the valve-circle, and then lay off 
this angle a from the dead center line OZ, in a direction opposite 
to the rotation of the crank, aud this will give the desired posi- 
tion of the eccentric. 

The angle ZOE = a is estimated from the portion of the line 
OZ to the right of center 0, and so long as the eccentric (or slide) 
is to the right of its middle position the angle made by the eccen- 
tric with the portion OZ of the reference line will be less than 90°. 
When this is the case the diameter OD of the slide-circle is always 
less than 90° from the crank position, which ensures that this circle 

{actual I 
1 J > crank whenever the eccentric or slide 

prolonged J 

is to the I ^J^^ \ of its middle position. This is true for all values 

of a, for all positions of the crank, for either right- or left-handed 
rotation, and whether slide-seat is to the right or left of shaft 0» 
provided, of course, that there is no reversing lever between 
eccentric and slide. 

The position of this valve-circle does not change while the 
crank rotates, its diameter making with the fixed reference line 
(9Zan angle DOZ that is equal to the constant angle COB be- 
tween crank and eccentric. That this is true is evident from : 

COE = COD -f DOE = ZOE -f DOE = ZOD. 

The valve-circle's position is therefore dependent only on the 
relative position of eccentric to crank. Provided this eccentric 
setting remains exactly the same in direction and magnitude, 
the crank may have either right-handed or left-handed rotation, 



RELATION OF VALVE-CIRCLE AND ECCENTRIC SETTING. I 39 

may occupy any position in any one of its quadrants and yet 
give the self-same valve-circle whenever the construction de- 
scribed above is strictly followed. 

For a given eccentric setting the valve-circle will have the 
same position, that is, the slide's distance, position and direction 
of motion relatively to its center of travel will always be the 
same for the same crank position. But when the eccentric setting 
is different, the motion of the slide relatively to the center of 
travel will be different, even though the crank position and rota- 
tion are exactly the same. This is illustrated in Fig. 6 1 : 

I. Slide is to the right of its middle and is going towards it. 
II. Slide is to the left of its middle and is going towards it. 

III. Slide is to the left of its middle and is going away from it. 

IV. Slide is to the right of its middle and is going away from it. 

The corresponding valve-circles are given in the second row of 
Fig. 6 1 when the crank position is OC and the rotation right- 
handed, and in the third row when crank position is OC and 
the rotation left-handed. The angle between the crank and 
eccentric being the same in magnitude for both rotations, the 
laying off angle <t, used in finding location of diameter of valve- 
circle, will be the same for the corresponding figures of the 
second and third row. 

In the last row of Fig. 6 1 we have the same four, possible, 
types of slide valves which are kinematically exactly like the 
corresponding Cases I, II, III, IV of the first row, and have the 
same valve-circles. The eccentric driving these slide valves is 
not shown because convenience of illustration requires the valve 
to be placed above the cylinder and because the setting of the 
eccentric relatively to crank is shown in the first row for both 
right- and left-handed rotation. Each valve is closing left port. 

Each of these four types has its special, valve, characteristics. 

A valve is said to have I ^ T > lap when the port is 

( negative J ^ ^ 

< /in the valve's mid-position. We will call a valve 



140 RELATION OF VALVE-CIRCLE AND ECCENTRIC SETTING. 




Fig. 61. 



^dge {indSctf when it closes the left port (or passage leading to 



left end of cylinder) by moving to the ] ^Jht} • When all edges 
of the valve are \,^S:.\ we will call the valve itself {i^a} • 
When some of the edges are direct and others indirect (as in 
the ordinary D valve where the steam edges are direct and the 
exhaust edges indirect), we will call the valve itself direct or in- 
direct according as the edges regulating admissio7i of steam are 
direct or indirect * These definitions hold whether the ports are 



* English and American engineers, so far as the writer is aware, have 
never used these terms, direct and indirect. WiEBE makes use of them in 
his "Darstellung der Verhaltnisse der Schieberbewegung." They are short, 
clear, terms and may profitably be employed in this extensive subject. 



TYPES OF VALVES. I4I 

in the seat or the face of the valve. In Fig. 6i, fourth row, we 
have 

I. A valve with positive lap and direct cut-off. 
II. A valve with positive lap and indirect cut-off. 

III. A valve with negative lap and direct cut-off. 

IV. A valve with negative lap and indirect cut-off. 

We will designate them briefly as, I positive direct, II positive 
indirect, III negative direct and IV negative indirect valves. 

The ordinary D valve in such extensive use is positive direct 
on its steam side and positive indirect on its exhaust side. The 
Meyer expansion valve with its halves placed so close together 
that in the middle position they fall within the outer edges of their 
own steam ports is an example of a negative direct valve, and 
the same expansion valve when its cut-off plates are placed so 
far apart that in the middle position of the valve its inner edges 
are outside of the inner edges of their own steam ports, is an 
example of the negative indirect valve. 

Other positions of these valves between these extremes will 
give positive direct and indirect types. With the same eccentric 
setting therefore it is possible for a slide to represent any one of 
the four types of valves, according to the position of the pair of 
ports relatively to the middle position of the slide. (But with 
the same eccentric setting and same throw of valve the cut-off 
will be different for each type.) The terms positive and nega- 
tive, direct and indirect, refer only to this relative position of 
ports to valve and are entirely independent of the ecce^itric setting. 
(When the valve controls only one port, instead of two, the terms 
direct and indirect disappear entirely, only the distinctions posi- 
tive and negative then remaining.) The angle between crank 
and eccentric, measured from the former in the direction of rota- 
tion, for any one of these valve types may vary from o° to 360°, 
although commonly each type has its eccentric set in the eccen- 
tric-setting-quadrant shown in Fig. 61. 

In all of the engines of the fourth row of Fig. 61, for the same 
rotation, the same cut-off is taking place during the same stroke. 

JO 



142 



SELECTION OF VALVE-CIRCLES. 





Fig. 62. 

Hence four different types of valves can be arranged to accomplish 
the self-same object. The eccentrics which drive them will have 
four corresponding settings relatively to the crank. There will be 
four different valve-circles to represent the motion. They are 
grouped together in the two diagrams of Fig. 62, which differ 
only in the direction of crank rotation. 

To find the valve-circle belonging to any type we draw the 
crank position OC corresponding to the cut-off effected by the 
valve, and prolong it beyond center 0. Then lay off the lap, 
Oa = Oc, and draw circle ehfg with the radius of eccentric. At 
a and c erect the perpendiculars hae and gcf and thus determine 
the ends e, /^,/and^ of the diameters Oe, Of, Og, and Oh of the 
four valve-circles. To find the one belonging to the type of valve 
under consideration, notice whether this valve, when at cut-off, is 
to the right or left of its middle position on its seat. If it is to 
the {"fft} one of the two circles -jJi/ni,} "^vill be the desired one, 
because these cut the \ proSnged \ crank. This now reduces the 
choice to one of two circles. Now notice whether the valve in 
its cut-off position is moving towards or away from its middle 
position. If it is moving { aw^y ffom } this mid-position the chord 

as the crank 
In Fig. 62 chord 



Oa (or Oc) of the proper valve-circle will 
continues the rotation from its cut-off position 



diminish 
increase 



PORT OPENINGS BY DIFFERENT TYPES OF VALVES. 



143 



Ob is accidentally equal to chord or lap Oa, but it is not so 
generally. 

Hitherto in valve-gear discussions two diametrically opposite 
valve-circles were employed for they helped to show the main 
occurrences in the steam distribution at both ends of the cylinder. 
But in designing valve-gears we usually pass from the valve- 
circle to the actual gear and there is then danger of confusion as 
to the eccentric setting when two circles are used. There can 
be no objection to the use of two circles provided they can be 
distinguished by some means, say, by drawing the representative 
one in full lines. We shall however make use of but one valve- 
circle when there is but one crank. When there are two cranks, 
diametrically opposite, as in the Westinghouse engine, we shall 
use two valve-circles. 

We will not here describe the occurrences of the steam distri- 
bution, as this is given in many of the elementary treatises on 
the steam engine and we have supposed the reader to be familiar 
with the elements of this subject. Besides, the occurences for 
the ordinary D valve are inscribed on the Zeuner valve diagram 
in Fig. 63. But the following table may be of assistance when 
some of the unfamiliar valve-types are under consideration. 



TABLE XIV. 

Periods During which Port Controlled by Valve is Open or 

Closed (Given by the Limiting and Intermediate 

Crank Positions). 







Type of Valve.* 1 


Condition of Port. 


I 


II 


III 


IV 


Positive 
Direct 


Positive 
Indirect 


Negative 
Direct. 


Negative 
Indirect. 


Right Port. 


Open 


3-4 


1-2 


2-3-4-1 


4-1-2-3 


Closed 


4-1-2-3 


2-3-4-1 


1-2 


3-4 


Left Port. 


Open 


1-2 


3-4 


4-1-2-3 


2-3-4-1 


Closed 


2-3-4-1 


4-1-2-3 


3-4 


1-2 



* Left port closed by positive direct valve when crank is at 2. 

" " " " " indirect " " " " 4. 

" " " " negative direct " " " " 3. 

" " " " " indirect " " " " l. 




144 RELATION OF PISTON TO VALVE-TRAVEL. 

This table is applicable to both exhaust and admission by 
considering the lap circles of the adjacent figures to represent in 
the former case the inside, exhaust, lap, and in the latter case the 
steam lap, provided account is taken of the valve type of the edge 
in question. Intercepts of lap and valve-circle measure openings. 

As the valve-circle gives us the valve travel to the right or left 
for any crank position and Tables, pp. 26-27, the corresponding 
piston-position, we can represent graphically by an oval sort of 
curve the desired relation between piston movement and valve- 
travel. The ordinates of this curve are the chords cut from the 
crank by the valve-circle and the abscissas are the distances of 
piston from end of stroke given by Tables V and VI. Lines 
drawn parallel to the base line, and above and below the latter a 
distance equal to the outside and inside laps, give the principal 
occurrences in the steam distribution. The diagram is easily 
drawn when the valve-circle has been found. 

There is another diagram which requires still less work and 
gives readily the actual piston and valve travel for any crank 
position. Let OA = OB, Fig. 63, equal the crank radius to 
some reduced scale. For convenience we will take the length 
of the connecting-rod three times that of crank. Then with the 
length CB = 3 OB of this rod describe the arc EBF and with 
the same radius strike off through A an equal and parallel arc 
GAH. The horizontal lines included between these arcs will of 
course be equal to the stroke, and the horizontal intercepts be- 
tween these arcs EF and GH a.nd the crank-pin circle AJLBU 
will represent exactly the distances of the piston from one or the 
other end of the stroke. For instance, when the crank-pin is aty, 
the intercept IJ will represent the distance s of the piston from 
the left end of stroke. This is evident from equation (20) p. 35, 
where 

s T=z R{\ — cos co) -}- Z (i — cos a) 

Here the first term of the second member is equal AZ =^ WJ 
and the second term of this member is equal to VA = IW. 



STEAM DISTRIBUTION SHOWN BV VALVE-CIRCLE 




Fig. 63. 



14^ PORT OPENINGS SHOWN BY DIAGRAM. 

When the crank is at the cut-off positions OL and OS, the 
piston travel is KL and RS for forward and return strokes re- 
spectively, and will be found to be unequal. The valve travels 
for the same instants are equal to Og ; the scale for valve travel 
will usually be different from that for piston travel. At an- 
other crank position, ON, the piston travel is given by MN and 
the valve travel by Od. 

When the crank or its prolongation traverses the \ ^^j l^^j \ 

area, one or the other of the steam ports is open for <^ ^ "^ssi \^ ^ 

When crank is at F, the distance of the piston from beginning 
of the stroke is HY, the valve travel is Os, the opening of the left 
steam port is us and is closing, while at the other end of the 
cylinder the steam is exhausting through a fully open port, pt. 
In Fig. 63 the four crank positions 1,2, 3, 4, are drawn which 
correspond to the opening or closing of the steam ports by the 
admission edges of the valve; the crank position i^ 2', 3^ 4S 
correspond to the opening or closing of the steam ports by the 
exhaust edges of valve. This diagram is for a positive valve. 
It will be a profitable exercise to closely compare the steam dis- 
tribution given by Table XIV with that given by Fig. 63. 

An ordinary D valve is usually set by giving it equal leads 
at the dead points of the crank ; but this makes every other 
occurrence of the steam distribution take place unequally for the 
two strokes. We have already pointed out that there is no 
necessity for equality of lead; that the lead may be unequal 
provided the minimum value is sufficient to maintain the steam 
pressure, and cushion properly ; nor is a positive lead necessary, 
there are cases when a negative lead is justifiable and desirable. 

Engine makers lay some stress upon having the indicator 
cards from the two cylinder ends look alike. Equality of cut-off 
has more influence on this than any other factor. For this rea- 
son and not because equality of cut-off is necessarily conducive 



ECCENTRIC SETTING FOR A GIVEN, EQUALIZED, CUT-OFF. 14/ 

to smooth running, we will give a method* for finding the un- 
equal laps and the angle between eccentric and crank when the 
cut-off is equalized for both strokes, say, made equal to 0.8. 

In Fig. 64 take the distance 00' equal to the sum ^ -f ^' of 
the laps at the two ends, and on stroke AB and A'B' describe 
crank-pin circles. Strike off the equal and parallel arcs BK d^nd 
A'R with a radius equal to the connecting-rod, as in Fig. 63. 
Make KL = RS = 0.8 X AB7 In the neighborhood of Z draw 
a series of perpendiculars to 00' which cut the lower part of 
crank-pin circle BLA at points u^ . . . u^. Then with 5 as a center, 




and the distances, Lm^, Lu^, Lii^, Lu^, etc., as radii, strike off the 
arcs that will cut the corresponding perpendiculars through u^, u^, 
«3, u^, etc., at the points x^, x^, x.^, x\. Join these intersections 
by a curve h. It will cut at J/ the other circle on AB'. 

From this point M drop another perpendicular MN to 00' 
cutting the latter in P and the lower part of circle ALB in N. 



* See Burmester's 
675 and 676. 



Lehrbuch der Kinematik," pp. (i'ji-6']\, and Figs. 



148 EQUALIZED CUT-OFF FOR GIVEN ECCENTRIC SETTING. 

Then will chord LN = SM by construction and the angle 
LON = angle SO'M. Each of these angles is equal to the 
angle between crank and eccentric and hence to the angle between 
diameter of valve-circle and dead point line of eccentric (see p. 
137). For this last angle DOB = DOL + LOB, Fig. 63, and 
the first part DOL has a cosine = lap -=- eccentricity. The sec- 
ond part LOB is represented by LOP ox SO'Pvcv Fig. 64 and the 
first part by PON and PO'M. n^ncQ PON -\- LOP = L^ON 
= PO'M -\- SO'P= SO'M= angle between crank and eccen- 
tric. By giving the greater lap OP to the end of the valve 
farthest from shaft and lap O'P to the other end of the valve the 
two cut-offs will each equal 0.8 and the eccentric setting = LON. 
The other intersections of auxiliary curve // with A' MB' are of 
no use in this connection. 

Dr. Burmester also solves another problem, namely, one in 
which, for a given setting of eccentric, the cut-off shall be equal. 
Construct circles on stroke AB and A' B' with same center and 
0' as before. Then take any point / in circle ALB and lay off 
angle /(9^ equal to given setting of eccentric. Through e draw 
€e' perpendicular to (96^', ii^tersecting circle A' MB' at e'. Lay 




ANGLE OF ADVANCE DEFINED. 149 

off e' O'f on this circle equal to the given setting and through 
/' and /draw parallels .^y and rr/to 00' till they cut the arcs 
A'R and BK. Then make z'x = zf\ this will give one point 
X of auxiliary curve h, ^y similar constructions other points 
of this curve h can be found. The auxiliary curve h cuts the 
circle A' MB' in some point ^. Finally make angle SO' M 
=zfOe, drop the perpendicular MPN on to 00' and make angle 

NOL = fOc. Then will cut-off ^, = ^, and (9/^ will be the 
-^ A'B' AB 

outside lap of the valve at the end farthest from the shaft and 
O'F the lap for the other end of the valve. 

In the mathematical discussions of valve-gears, it is found 
more convenient to use the angle of advance d instead of the 
eccentric setting or angle between crank and eccentric. It may 
be expressed by a formula or it may be measured in any valve- 
gear as follows : Start with eccentric arm half-way between its 
own dead points and then move the crank to its (the crank's) 
nearest dead center. During this motion the angle passed 
through by either crank or eccentric will be the angle of advance 
and it will be positive if this motion or rotation is in the same 
direction as the engine rotation, otherwise it will be negative. 
The formula for this angle at any crank position is : 

S =^ X^oj' — 90°* {S6) 

Letting x represent angle between crank and eccentric, we have 

X = go° -{- d ±: x = ^^ — co' ± x> i^7) 

where y represents the angle between the two dead point lines and 

* According to Rankine : " By angular advance is to be understood the 
angle at which the eccentric arm stands in advance of that position, which 
would bring the slide-valve to mid-stroke when the crank is at its dead 
points." A represents the angle made by the eccentric with its own dead 
point, w' the angle made by the crank with its dead point line ; either of these 
angles may be greater than 180^. 



150 LEVERS BETWEEN ECCENTRIC AND VALVE. 

I lower I ^^^^ ^^ "^^^ when dead point line of eccentric 
] orecedes [ ^^^^ ^^ ^^^ crank in the direction of rotation. 

When there is a reversing lever between eccentric and valve the 
formula for the angle between crank and eccentric then becomes 

x' = 180° —x = go° — d :+^ X' (88) 

Sometimes a reducing lever is placed between the valve and 
eccentric for the purpose of reducing the size of the latter and 
thus diminishing its tendency to heat under high speed. The 
only modification which this makes in our diagram is to make 
the valve-circle diameter correspondingly larger. For we as- 
sume that the valve is driven directly by the eccentric. Some- 
times, as in Corliss valve-gear, a reducing lever is used to give a 
sort of differential motion, and the valve direction and chord of 
lever-pin are made to differ purposely.* 

The introduction of a reversing lever between eccentric and 
valve has the same effect on the valve motion, as if the eccentric 
were shifted to a diametrically opposite position and the valve 
then driven directly (i. e. without reversal) by the new eccentric. 
Hence here also, unless the contrary is specified, it is to be 
understood that diagram is drawn as if valve were driven directly 
by the eccentric. 

This finishes our discussion of the simplest case of valve 
gearing. Under the head of single-valve gearing, with invariable 
steam distribution, there still remains a case which is mainly of 
interest because of its bearing on link motions, namely, the case 
in which the valve stroke does not pass through the center of 
the shaft, but at a certain distance c from the latter. In this 
mechanism the stroke of slide is more than twice the length of 
the eccentricity ; moreover the two dead points of the crank are 
not diametrically opposite as in the ordinary slider-crank. 

* See also levers 0' RP and O' RQ! Fig. 82, Porter- Allen Engine. 



VALVE TRAVEL IN CROSSED SLIDER-CRANK. 



51 



In Fig. 66 F is a valve, whose line of stroke VV passes 
center of shaft at distance OV ^ c. It is driven by an eccentric 
6^^ r=r and rod ^F of length /. In valve-gears the point from 
which valve-travel is usually estimated is one near the middle of 
the valve-stroke ; the point of valve whose travel is measured 
may be any point rigidly attached to valve ; in this figure the 
right-hand end Fof eccentric-rod is the one chosen. If we find 
the position of V corresponding to each of the crank's dead 
points and then bisect the distance between these two positions 
of V, the point of bisection will be the point from which valve- 
travel is measured. 

Let Mhe this center of valve-travel and OE any position of 
the eccentric. MVwiW then be the corresponding valve-travel 
or distance from middle position. At the point E draw EC equal 




Fig. 66. 

and parallel to MV, and join MC. The quadrilateral CEVM'xs 
consequently a parallelogram and CM -- EV ^= I. In other 
words, the locus of the point Cis a circle described from J/ as a 
center, with radius /. The valve-travel for any other eccentric 
position OE' can therefore be found by drawing the parallel E'F 
up to the arc GDHf^^ Draw the chord GOH of this arc per- 



"^ The travel of the valve from the end of the stroke for any position OE' 
of eccentric is evidently either E^ S or E' R, according as one or the other end 
of the stroke is meant. The arcs MRO and NSP are struck, from the ends 
A and B of valve-stroke, with radius / of rod. 



152 VALVE TRAVEL IN CROSSED SLIDER-CRANK. 

pendicular to line 0M\ in this figure arc and chord meet at 
points G and H that seem to be on eccentric circle EE'R ; but 
they do not necessarily lie on this circle. 

Even in this sort of valve gears the rod / is long in compari- 
son with the radius r, the rise OD of the arc is very small and 
the arc may then be replaced by the chord GH passing through 
center 0. The distance EI, parallel to stroke, now represents the 
valve-travel, and for any other eccentric position OE' the dis- 
tance from E' to the straight line GOH measures the travel of 
the valve. When eccentric center E and point H are on the same 
horizontal, the valve-travel EH will be the same wheather mea- 
sured up to chord GOH or arc GDH\ the same may be said 
when E and point G are on the same parallel to stroke. These 
two points of E, if constructed, would be found to be diametri- 
cally opposite and equidistant from arc or line of reference ; if 
from each of these ^'.f as a center and with radius / we describe 
an arc, cutting stroke AB, and then bisect the distance between 
these arcs the point of bisection will be the center of reference M 
assumed above. 

But the travel may be more easily found and this case reduced 
to that of the ordinary eccentric, by finding an eccentric Oe 
whose distance ^^from the vertical 6^F is always equal to EI^. 
To find such an eccentric, we drop from point /a perpendicular 
IL upon the eccentric OE, and erect Ee perpendicular to OE. 
Now from the point K, where H cuts the vertical F, we draw 
Ke parallel to ^/and join and e. It is evident that eEIKis a 
parallelogram and eK = EI. To be an equivalent eccentric Oe 
must not change its length or position relatively to OE, that is, 
Ee = //Tmust be a constant for all positions of the original 
eccentric OE. The figure shows that triangle OKL is similar to 



* Up to this point the demonstration is like that given by Prof. A. Fliegncr 
in his work " Umsteuerungen der Locomotiven." 



SINGLE-VALVE GEARS, STEAM DISTRIBUTION VARIABLE. I 53 

each of the triangles IJK and OJE. These triangles are there- 
fore similar to each other, and we have 

— — r= -^ = tangent of angle HO Y= - nearly. (89) 
OJz OJ I 

But this angle is constant, being equal to the angle MO V\ 
hence IK = Ee is 3. constant, and Oe has all the properties of a 
virtual or equivalent eccentric. If we connect ^ by a rod / 
with a slide V, whose stroke passes through 0, the travel 
of V from its middle position will be very nearly like that of 
the slide V from M. Evidently stroke of F is greater than if 
driven by B. 

The distance of -fi'from chord GOH, or of e from vertical OY, 
does not exactly represent the travel of the valve, for it does not 
take account of the angularity of the eccentric rod (though it 
does consider the direction of line OM, and thus takes partial 
account of the length of this rod when distance c is given). The 
exact valve-travel is only given by the distance of eccentric- 
center E from arc GDH. 



Single- Valve Gears, Steam Distribution Variable. 

The characteristic feature of a single-valve gear is that its 
valve slides on a fixed, stationary seat (p. 134). In this sense an 
engine may contain one or more single-valve gears which divide 
among themselves the functions of the steam distribution. 

The variable elements in these gears exist mainly to vary the ex- 
pansion. They may roughly be arranged into two groups, those 
in which the driving eccentrics themselves can be varied, and 
those in which the eccentrics are themselves non-adjustable, the 
variation in the valve motion being effected by mechanism be- 
tween the eccentrics and the valves. To the second group belong 
the link-motions, and these we will take up in the Appendix, 
because they are more complex and less extensively used in 
high-speed engine work than the first group. 



154 VALVE-GEAR VARIATIONS. 

In the first group the eccentric is varied by simultaneously- 
changing both its throw and angles of advance * The principal 
means by which this change is accomplished is the *' swinging 
eccentric " device. In this the center of the eccentric is moved 
across the shaft in an arc LL having the point Pas its center, as 
in Figs. 68 to 71 and Figs. 73 to yZ. Ten of the " Single- Valve 
Automatics " represented use this device, the mechanisms for 
producing the swinging motion differing more or less ; (see cata- 
logues of engine builders). The object of this change is to alter 
the power of the engine by altering the amount of expansion, 
but other changes in the steam distribution also take place at the 
same time when there is but a single valve to effect them. For 
example, the compression begins earlier when the expansion 
begins earlier, great cushioning going hand in hand with great 
expansion. These two functions of the steam distribution have 
perhaps the most influence on the form of the indicator card and 
thus are the principal means of regulating the power per stroke 
and the uniformity of the driving force. These are important 



* The only engine known to the writer that uses a single valve and varies 
the cut-off by varying only the angle of advance, is the oscillating engine 
built by the J. T. Case Co. Strictly speaking this engine has two valves, 
for the ports in the rocking cylinder control release, exhaust closure and 
admission, the cylinder itself performing the functions of the main valve. 

On the other hand, the Buckeye engine has apparently a double-valve 
gear and yet in reality is made up of two single-valve gears the first being 
the one to which the main valve belongs and the second having the expan- 
sion valve. In this second gear, the expansion valve slides on a moving 
main valve, but the mechanism is such that the relative motion of expan- 
sion to main valve is scarcely affected by the absolute motion of main valve. 
The latter therefore acts as a stationary seat of the expansion valve which 
may be discussed and designed as if the ports it controlled were fixed and 
as if its rocker arm had a stationary pivot. We have however discussed 
the Buckeye's gear in Figs. 96-98, with the double-valve gears, vi^hich it so 
strongly resembles externally. When an engine has its distribution ef- 
fected by several valves, each sliding on a fixed seat, the functions of one 
of them may well be varied by varying only the angle of advance of its 
eccentric or only the throw or lap. 



EQUALIZATION OF THE DISTRIBUTION. I55 

factors in securing steady running of the engine. To effect 
smooth running there must be not only absence of great fluctua- 
tions of power per revolution, but also absence of vibrations and 
shocks. Shocks occur when there is " play " (or clearance) at a 
bearing accompanied by such sudden and complete changes of 
direction in the force acting on the bearing that continuous con- 
tact between the pin (or slide) and its bearing ceases while the 
space (or " play ") between the two is traversed by one or the other 
of the two pieces. Tremors and vibrations of course accompany 
shocks, but they may also arise independently, from sudden 
changes in the intensity of the pressure between two pieces. 
Thus at the end of a piston-stroke the final cushion-pressure may 
be considerably less than the initial pressure on piston at begin- 
ning of next stroke, the sudden, additional load on piston send- 
ing a tremor through all the connections. Sudden reversals of 
pressure should therefore be avoided particularly at the dead point 
where they are most dangerous. As the extent to which 
cushioning is carried, the character of the lead (positive or neg- 
ative) and its extent, strongly influence these dead-point-pres- 
sures, the variations effected by the valve gear in these quan- 
tities per revolution and at the different grades of expansion 
become a matter of consequence. In times past engineers laid 
great stress on having the steam lead equal at the two cylinder 
ends whatever the grade of expansion, particular care being taken 
to " set " the valve with equal lead. This desire for equalization 
at the two cylinder ends extended itself to the cut-off, the release 
and the cushioning as well as to the lead. But it was recognized, 
particularly in locomotive practice, that it was impossible with 
existing valve gear to equalize all the functions at the same time, 
equalization of any one function was at the expense of the others. 
Generally it was the lead (a positive one) that was thus favored. 
At the same time most engineers also held that a lead that was 
constant at all grades of expansion was a desideratum. It was 
regarded as the special merit of the Gooch link-motion that it 



6 VARIATION OF THE STEAM LEAD. 



possessed this virtue. But since the advent of the high-speed 
steam engine a difference of opinion on this matter has developed 
itself among engineers. 

In the pioneer engine of this type, the Porter- Allen, the leads 
were deliberately made unequal at the two ends (Figs. S6, 8/ and 
Figs. 89, 90 of Porter-Allen valve diagrams) and the lead also 
varied from mid-gear to full gear, as in other link-motions. The 
Straight Line engine in its earliest forms had a lead that was 
equalized at the ends and was nearly constant at all grades of 
expansion ; in the later forms the lead is now variable for dif- 
ferent cut-offs ; it is sometimes negative and there is some in- 
equality at the ends. In other and excellent engines the old views 
on this subject are still carried out so that the matter cannot be 
regarded as finally settled either way, Prof Sweet, the designer 
of the " Straight Line " engine, going so far as to determine ex- 
perimentally, with the indicator, the conditions of smooth run- 
ning for each engine. The collection of examples given shows 
the variation of American practice in this respect. We shall not 
attempt any comparisons of these engines, for the exact effect of 
the steam distribution on the steadiness and smoothness of run- 
ning can only be determined by a knowledge of the clearances, 
the pressures, the weights of the reciprocating parts and the reg- 
ulating capacity of the governor under variable loads. 

We have thus far said nothing concerning the release, because 
it has usually less influence on smoothness of running than the 
other factors, though it too effects somewhat the pressure on 
piston at dead point. The release is of consequence to econom- 
ical running because of its influence on the amount of back 
pressure, an early release and widely opened exhaust ports tend- 
ing to keep this pressure small. On the other hand release 
begins too early when it curtails the period of expansion to any 
notable degree. We have given the beginning of the release for 
only a few cases ; it may be easily found from the exact dia- 
grams, by following the directions given with Fig. 6y. 



SHAPE AND POSITION OF LOCUS. 15/ 

Returning now to the influence that the shape and location of 
the locus LL of the eccentric centers E has on the steam distri- 
bution, we note first, that as the chord of the utilized portion of 
the arc LL is more nearly perpendicular to the dead point line 
of the eccentric, the lead is more nearly constant at the different 
grades of expansion, provided the curvature is the same, and 
secondly, that with this perpendicular location an arc LL of 
great radius (i.e., of slight curvature) causes smaller variations in 
the lead than an arc described with a small radius. 

Radii drawn from the center to the locus LL (or E^E^E^ of 
centers, give the length and position of the eccentrics corre- 
sponding to the different grades of expansion, the locus moving 
with the crank to which it is rigidly attached. The broken line 
DqD^D^ is the locus of the vertices of the diameters of the valve- 
circles corresponding to the eccentrics E^.E^.E^. The position 
of these diameters were found from their eccentric by laying off 
the angle a as in Fig. 60. In Fig. 74 the locus L'L' or E^^E^^E^^ 
was derived from LL, the locus actually described by the swing- 
ing eccentric, by constructing a series of offsets each equal to 

--^ r^ , according to the method given in connection with Fig. 66. 
34 

The new eccentrics thus obtained are regarded as virtual eccen- 
trics and their centers are indicated by E^^^MviA^- The valve 
circles were derived from the virtual eccentrics OEv^, OE^^, OE^^- 

The particular eccentrics chosen for representation are those 
corresponding to minimum, quarter and maximum cut-off, the 
first and last of these being taken from data furnished, in most 
cases, by the manufacturers themselves. The quarter cut-off is 
generally an average value of the cut-offs in the two cylinder 
ends when angularity of connecting-rod is taken into account. 
The intermediate cut-offs in the case of the Westinghouse 
engines are somewhat larger than 0.25 or about 0.30. Each o^ 
these valve-circles will give the steam distribution shown in Fig. 
63, when the lap and port- width circles are drawn. The crank 
II 



158 ANGULARITY OF ECCENTRIC ROD. 

positions and piston travel at cut-off, preadmission and beginning- 
of compression are only drawn for the eccentric and valve-circle 
corresponding to % cut-off; the shaded area shown in the diagram 
cuts, from the crank, intercepts that represent the port-openings 
at quarter cut-off. If there is more than one port each intercept 
must be multiplied by the number of ports to get the total 
opening. It must not be forgotten that each of these valve- 
circles cuts from the crank or its prolongation a chord that 
represents the distance of the eccentric center E from the perpen- 
dicular YY, through (9, to the dead center line of the eccentric. 
With the exception of Figs. 74, 77, this perpendicular is the vertical 
through O. The valve-circles entirely neglect the angularity of 
the eccentric rod. Whenever the length of the latter was known, 
a dotted, central, arc of reference was drawn from which the 
valve travel from the center of this arc can be found exactly. 
This arc center is obtained as follows : the crank is placed at one 
of the dead points and the center E of the eccentric in question 
(usually the one for maximum cut-off) is placed in its corres- 
ponding position, then with ^' as a center and length of eccen- 
tric-rod as a radius strike off an arc cutting the valve-stroke ; 
repeat the operation for the other dead center of crank, the 
point bisecting the distance between these arcs will be the center 
of the arc of reference required. In the Rice, Ball, Southwark, 
Straight Line and Westinghouse compound engines, the central 
arc of reference differs so little from the vertical through that 
no effort was made to draw the polar curves representing exactly 
the valve-travel, the valve-circle being sufficiently accurate for 
all practical purposes. In the Westinghouse Standard there is 
also only a slight difference between the perpendicular YY' and 
the arc Y' Y' ; still for the case of maximum eccentricity the 
exact polar curves were plotted on the diagram in dotted lines, 
the one outside the valve-circle corresponding to the positions 
of the valve to the left of the center of the reference arc and the 
one inside to the positions of the valve to the right of this center. 
Comparison with the intermediate valve-circle shows that the 



SYMBOLS USED ON DIAGRAMS. 159 

steam distribution is but slightly changed by substituting the 
exact polar curves for the less exact circle. 

The cut-offs are apparent ones and are estimated from begin- 
ning of stroke in the same manner as in former sections. Pre- 
admission, beginning of compression and release are however 
measured from the end of the stroke and each is then divided by 
the stroke itself, so that the results are always expressed as 
fractions of the stroke. The release is not tabulated on the 
diagrams. The symbols used are 

a = width of port. 

/ = length of port in cylinder. 

d = diameter of piston valve. ^ 

e = steam lap. 

i = exhaust lap. 

£ z=z apparent cut-off. 

V = steam lead. 

r = eccentricity. 

R = radius of crank. 

L = length of connecting-rod. 

£ =z beginning of release, measured from end of stroke and 

divided by stroke. 
7^ = beginning of compression, measured from end of stroke 

and divided by stroke. 
■// = beginning of admission, measured from end of stroke 

and divided by stroke. 

The period of compression would properly be represented by 
the difference rj — r/. 

The subscripts o, i and ^ affixed to these symbols signify that 
they belong to minimum, quarter and maximum cut-offs respec- 
tively, as explained above. 

All the engines represented "run over" (see p. ii8), i. e., in 
these diagrams, the rotation is right-handed. Then in Fig. 67, 
for the distribution corresponding to quarter cut-off, ATT" repre- 



l6o SINGLE VALVE GEARS. 

sents admission of steam to head end during forward stroke and 
SR to crank end for return stroke. FIf represents compression 
n crank end during forward stroke and 6^ £/ compression in head 
end during return stroke because exhaust lap is negative, pi q^ 
represents the preadmission to head end before beginning of 
forward stroke and pq the preadmission to crank end of cylinder 
for return stroke. As the inside lap at crank end is different 
from that at head end we must carefully distinguish between the 
two. Then m^ni is that last part of the forward stroke during 
which release takes place in the head end and mn that part of 
the return stroke during which release takes place in the crank 
end of the cylinder. By dividing these quantities by 2 X CO, 
we get the values tabulated in the diagrams. We have omitted 
the release from these tabulations but it can be easily found from 
these exact diagrams if desired. In following the steam distri- 
bution by means of the diagram, it is well to bear in mind that 
the valve is to the right of its middle position whenever the 
valve-circle cuts the crank radius. 

The same letters are used in the other diagrams to represent 
the same parts of the distribution ; the letter C always represents 
the crank position, E^ eccentric position corresponding to C^ and 
Dn the diameter of the valve-circle corresponding to the eccen- 
tric-setting COE„. 

The valve diagrams, for the twelve Single-valve Automatics 
represented, are here divided into two groups according to the 
location of the pivot P (or center of curvature) relatively to the 
crank radius. In the first group. Fig. 67-72, the center of cur- 
vature of that element of the locus LL which crosses the crank 
lies on the crank's own radius, while in the second group. Figs. 
73-78, the pivot P lies outside of the crank's radius or its pro- 
longation. As the pivots P are often at considerable distance 
from their loci LL, economy of space required that the points 
be detached from their proper position relatively to their loci 
and be placed nearer the latter ; this has been done in every case 
and their co-ordinates given so that they can be laid out if desired . 



LL 



1 



/ 



M 



/ 



/ 



\ 
\ 



Pigs. 77 and 78. 



WrSTINCHOUSE STANDARD VERTICAL 



91 X 9 



NDJRECT Valve. 




^EAO cTnTER line 

C"- 



S 






RIGHT CYLINDER LEfT CYLIMDERl RIGHT CYL. 

i = 0.023 0.023 71= 0.63i 
Oi Jo, 

^ = 0.283 0.283 7, = 0.37c 
f =0.75 0.75 n=0.\65 



LEFT CYU 

0.035 
0.0083 0.0083 
0.0025 0.0025 



RIGHT CYLINDER 

0.63! 1 7^=0.035 
0.373 Yl 

o.ieb ^ 

y 

WESTINCHOUSE CO!Vlt>OUND VERTICAL 

= 2 14 and 24x14 \ ENCTNE. Positive \ I iMDiRECT^ALVE ° 




GIDDINGS VALVE. 



i6i 



Russell & Co. Engine, 14x20. 

Fig. 6^. In this valve gear the center of the eccentric is 
moved in a perfectly straight line LL across the shaft, at right 
angles to the crank. The mechanism for accomplishing this (it 
is not the " swinging eccentric " device) is not given here but 
can be found in catalogue of the builder. Inspection of the 
valve-circles shows at once that the steam lead v is constant at 
all grades of expansion. The lead on the exhaust side is how- 
ever unequal at the two cylinder ends, for the inside lap 2 = 
and — 5^ at the crank and head end respectively. But this is 
not a matter of any consequence. The inequality in the inside 
lap equalizes the beginning of release and of compression for the 
two cylinder ends as the following table shows : 



£0' 


Crank End. 


Head End. 




Crank End. 


Head End. 


0.54 
0.36 
0.075 


0.54 

0.37 
0.075 


^0 

7]. 


0.458 
0.292 
0.065 


0.450 
0.284 
0.050 



The value of rj includes the preadmission of steam and so does 
not exactly represent compression proper. But it may be taken 
as representing the cushioning, which is practically equalized for 
the two ends by this arrangement. The table connected with 
the diagram gives the variation of cut-off and preadmission for 
minimum, average and maximum throw ; the range of cut-off 
varies from jV to t*^. This engine uses the Giddings' Balanced 
Valve, which is divided into two parts, rigidly connected, one 
part for each end of the cylinder. Each part admits steam 
through two passages that run through the valve and somewhat 
resemble the supplementary passage of the well-known Allen 
valve. Like the latter this valve belongs to the positive direct 
type and increases the clearance volume during the compression 
period by the volume of the supplementary passage. The clear- 
ance volume is smaller at the beginning than at the end of the 



RUSSELL «. CO., ENCINE. 



PAYNE ENCINE 

Positive 



BALL ENCINE.-OLDER TYPE. 

9 X 12 ^, Positive Indirect Valve. 



SOUTHWARK ENCINE. 

1 Positive Direct Valve.. 



STURTEVANT ENCINE. 

8 X 12 Y Positive Indirect Valve. 



WESTINCHOUSE STANDARD VERTICAL 

SJxS --'^^--r ^ENGINE. Positives Indirect Valve. 




1 62 



N. Y. S. S. POWER CO., AND PAYNE ENGINES. 



compression. As there are two ports or passages for each end of 
the cylinder, the admission and cut-off are prompt, and the 
port opening given by the shaded area of the diagram must be 
multiplied by 2. The valve is of the flat and balanced variety. 
The speed of this size of engine is 210 revolutions per minute. 

N. Y. S. S. Power Co., Engine, 12x12. 
Fig. 6S. The locus LL of the center of eccentric is concave 
to the shaft 0. The center of this arc is on crank radius, at 
pivot P, and 14'' from center of shaft. The lead diminishes 
as the cut-off increases. The cut-off ranges from ^j to f. The 
beginning of release is given by this table, 





Crank End. 


Head End. 




0.54 

0.35 
0.065 


0.458 
0.276 
0.046 



Payne Engine, 13x12. 

Fig. 69. The location of the pivot of the pendulum arm of 
the swinging eccentric in the smallest engines is different from 
that here given. In a 5x7 for instance, the pivot P and crank- 
pin C are on opposite sides of center ; in this case pivot P and 
pin C are on the same side of 0. The pivot is 1 1 }4" fron^ shaft 
center and lies on the same radial line as crank. The radius of 
arc LL is 10^''; this makes the arc convex to the shaft and the 
lead increases with the cut-off, the latter ranging from o to ^. 
The fraction of piston stroke occupied by preadmission is very 
small. The valve takes steam on the inside and is of the flat, 
balanced variety. This size of engine makes about 250 revolu- 
tions per minute. 

Rice Engine, 7x10. 

Fig. 70. This engine presents a unique appearance in that no 
governor is visible. The regulator is within the crank disc and 



RICE AND BALL ENGINES. 1 63 

turns an arbor that passes through the center of the crank-pin C. 
This arbor is the pivot P of the pendulum arm that swings the 
eccentric-pin across the shaft when the expansion is to be varied. 
The eccentric-pin and the pendulum arm constitute a sort of 
return crank. In the figure, both ^'and /^represent the center 
of pivot and of crank-pin. The locus LL is again convex to 
the shaft and has a nearly constant lead ; its radius is 41^. The 
fraction of the stroke occupied by the preadmission is larger in this 
engine than in any other of the dozen represented. The cut-off 
ranges from y to ttt. The length of the eccentric-rod is only 
\^" , but the influence of its angularity on the valve-travel is but 
slight, as the arc VY' shows. This arc was struck from the 
center of motion corresponding to maximum throw and with a 
radius of 15''. (The center of motion of valve is half way 
between the two valve positions corresponding to the crank's 
two dead points.) The vectors of the valve-circle represent the 
distance of point E^ from vertical FFand are very nearly equal 
to the valve-travel from the center of motion. The exact valve- 
travel at any instant is given by the distance of point E^ from 
the arc Y'Y. These exact travels could be laid off on the crank 
positions or their prolongations and thus two exact, polar, diagrams 
be found (for each eccentric throw) which would cut from the 
crank the exact distance of the valve from the middle position. 
But in this case they would differ so little from the intermediate 
valve-circle that this nicety of construction has been omitted. 
The valve is of the flat, balanced variety. 

Ball Engine — Older Type, 9x12. 

Fig. 71. The diagram of this engine is given mainly for com- 
parison with the diagram of a more recent style of valve-gear 
for the same engine whose diagram is given in Fig. ^6. In this 
earlier form the point P oi the swinging eccentric was 13^'' dis- 
tant from the center of the shaft and was on the same radial 
line as the crank-pin C. The locus LL of the center of the 
eccentric was convex to shaft and gave a lead that increased 



164 ARMINGTON & SIMS ENGINE, 

as the cut-off increased. The range of cut-off was from o to 5^. 
The valve was flat and balanced. The length of the eccentric- 
rod was about 41^ inches and with this as a radius an arc VY' 
is struck off from the center of motion, in the manner already- 
explained when discussing the Rice Engine ; the distance of the 
center £„ of eccentric from dotted arc Y^Y gives the exact 
travel of the valve, while the valve-circle gives the distance of £„ 
from vertical YOY; here again the difterence is so slight that we 
may rest well content with the results furnished by the valve- 
circles. 



Armington & Sims Engine, 14^x15. 

Fig. 72. In this engine a peculiar linkage is employed to vary 
the position of the center ^ of the eccentric; for the details of 
this linkage we must refer the reader to the catalogue of the 
engine builder. The locus LL of this diagram is not a circular 
arc, though it closely resembles some of the loci that have 
already been given. It was plotted from an accurately measured 
linkage belonging to a 14^^ x 15 engine. Here also, the lead 
increases as the cut-off increases, the table on the diagram show- 
ing the extent to which this takes place. The range of cut-off 
is from o to I. As in all the preceding cases the compression 
or cushioning period increases with the period of expansion. 
The eccentric-rod which connects the driving eccentric with the 
10^" rocker arm is itself 44" long. We may regard the rocker 
arc as a straight line and repeat the construction of the arc F' F 
already given with the Rice & Ball engines. Here also the dif- 
ference between vertical reference line FFand the reference arc 
P P is so slight that it may be neglected. The valve is of the 
hollow piston variety, is 6j{'' in diameter, is of the positive indi- 
rect type and takes steam simultaneously at two places. One of 
its steam passages leads through the center of the valve and in 
principle is like the supplementary passage of the Allen valve 
that is still used on some locomotives. This size of engine is 
run at about 280 revolutions per minute. 



SOUTHWARK AND STRAIGHT LINE ENGINES. 1 65 



South WARK Engine, 9x10. 

Fig- 73. It is the first of the second group of Single-valve 
Automatics. In this group the locus LL of the driving-pin or 
eccentric E of the valve has its center P outside of the crank 
radius OC. In most of these cases the pivot P and crank-pin C 
are still both on same side of shaft center 0. In the present case 
the co-ordinates of Pare 5 Y^" and — Y^" which brings it opposite 
the middle of the utilized portion of the arc LL The arc is 
concave to and its chord is perpendicular to crank radius so 
that it gives a nearly constant lead, for all grades of expansion. 
The cut-off ranges from o to ^ and the cushioning from xV to y. 
This valve-gear and its governor were designed by Prof C. B. 
Richards. The substitution of a pin E in place of the eccentric 
sheave is an excellent feature as it removes the danger of over- 
heating to which the eccentrics of high-speed engines are so 
liable. 

The rod covering the driving-pin E with the zVa-" rocker is 
34'' long; treating the rocker arc as approximately a straight 
line, we can repeat the constructions of the arc Y'Y' already 
given with the three preceding engines. Here too the deviation 
of the more exact reference arc V V from the vertical reference 
line FFis so Hght that there is no need of substituting exact 
polar curves for the valve-circles. The valve used in this engine 
is flat, balanced, and of the positive direct type. The speed of 
this 9x10 engine is 300 revolutions per minute. 

The Straight Line Engine, 11x14. 

Fig. 74. The locus LL is concave to and has the point P 
outside of the crank radius OC, but P and C are both on the 
same side of 0, the abscissa of P being 12.25'' ^.nd its ordinate 
— 2Y2" ■ The stroke of the valve does not pass through the center 
O of the shaft, but passes it at a distance of i Y2 inches. The dead 
ooint line of eccentric will therefore be different for the two ends of 



1 66 STRAIGHT LINE ENGINE. 

the valve strokes and for the different throws of the valve. An 
average position has been drawn on the diagram. It is therefore 
like the case represented in Fig. 66, p. 151. The offset Ee is 

there shown to be nearly equal to — r; here c =^ 1.5", / = 34'' 

and r = eccentricity corresponding to grade of expansion con- 
sidered. If r = 2.5'^ then -^ X 2.5 = o.ii is the offset, which 
laid off from E^ at right-angles to OE^ gives E^ ^ which is the 
center of the virtual eccentric OE^ ^, In Hke manner other offsets 
may be found for other throws of the valve and thus a curve L'L', 
passing through the ends of these offsets, can be found which will 
be the locus of the centers E.^ of the virtual eccentrics OE^. The 
valve-circle belonging to any eccentric can be found as in Fig. 60 
and in either of two ways : it can be obtained from the actual 
eccentric (drawn from to locus LL) by laying off the angle a 
from the dead point line of eccentric, or it can be obtained from 
the virtual eccentric OEy by laying off an angle o from the dead 
point line of crank. (See p. 137, Fig. 60 and Fig. 66, p. 151). 
The chord cut from the crank by the valve-circle thus determined 
equally represents the horizontal distance of En from the YY 
line through or the horizontal distance oi E^^ from the vertical 
line Y' Y'. (The chords of the valve-circle do not represent either 
of these distances exactly on account of an approximation made 
in the construction. Fig. 66) In neither of these determinations is 
the angularity of the eccentric-rod fully taken into account. To 
do this we proceed as in preceding cases. Thus from E^ as a 
center, with length of eccentric-rod (= 34'') as a radius we strike 
an arc that cuts the central line or stroke of the valve ; with 
same radius and EJ as a center a second arc is struck off, again 
cutting the Hne of stroke of the valve ; the point on the stroke 
bisecting the distance between these arcs will be the center of 
motion, from which as a center and 34'' as a radius the arc of 
reference Y"Y'' can be described. The distance of En from 
this arc gives with great exactness the actual travel of the valve 
from the center of motion for the assumed eccentric 0E„. la 



STRAIGHT LINE ENGINE. 



167 



this case, the arc of reference P^y' does not hug the line FF so 
closely that it is evident that the vectors of the valve-circles 
represent with sufficient accuracy the travel of the valve. We 
therefore give in the following table the exact steam distribution 
for ^2 = 2.5 and r^ = 1.48, the latter value differing a little from 
the value r^ = 1.41 assumed in diagram. The following values 
were found by constructing the exact polar curves. 



■I c 

is 


r^ = 


1.48. 


1 


^2 = 


2.50. 


Qi 


Crank End. 


Head End. 


Q.g 


Crank End, 


Head End. 


£1 


0.31 


0.29 


£. 


0.71 


0.77 


£x' 


O.IO 


0.12 


^'. 


0.04 


0.04 


^; 


0.34 


0.34 


7). 


O.II 


0.13 


V^ 


-\- 0.00 


— 0.005 


+ 0.00 


+ 0.00 



This is almost perfect equalization of the distribution for the 
two ends. A careful comparison of these results with those 
given in the diagram show that, with perhaps the exception of 
cut-offs near i^, the distribution is given with sufficient accuracy 
by the valve-circles even in this extreme case. 

The numerical values inscribed on the diagram show that 
cut-off ranges from about o to ^, that the cushioning is nearly 
equal for the two ends of the cylinder and that the lead varies 
from a negative value of o.ii at minimum throw to positive 
value of O.II at maximum throw. The lead is known to be 
about equal at the two ends for this throw. But this preadmis- 
sion is attained in very small fractions of the stroke as the tabu- 
lated values r/ show. 

The minimum throw chosen was that one in which the port 
is opened only for an instant. The almost perfect equalization 
of the functions of the exhaust illustrates the value for this pur- 
pose of making the laps unequal. There seems to be more 
clearance at head end than at crank end in this engine. 



1 68 STRAIGHT LINE ENGINE. 

Steam is admitted at two places, the port openings given 
by shaded area of the diagram must therefore be doubled in 
order to get the true area available for the passage of the steam. 
The speed of this size of engine is about 275 revolutions per 
minute. 

The Straight Line Engine was designed by Prof John E. 
Sweet and constitutes a most instructive example in steam distri- 
bution. I am indebted to Prof Sweet for the following outline 
of his practice in securing smooth running engines. 

Originally the valve motion was designed to produce a practi- 
cally constant lead at each end of the cylinder. Later experience 
convinced Prof Sweet that a constant lead was exactly what he 
did not want and he therefore located the eccentric plate pivot P 
so that it would give a variable lead equally at the two ends 
when a correcting, rocker arm, device was used. Still later he 
found that the variable lead enabled him to dispense with the 
correcting arrangement, the lead at the two ends being then 
somewhat unlike. Prof Sweet's practice on the testing floor is 
as follows : the first valve is made too long and with too much 
inside lap, then with varying loads and the indicator and the 
exercise of judgment as to smooth running, the valve is altered 
and tried until the best average is obtained. So much discrepancy 
has been found between different size engines of the same style 
and sometimes different engines of the same size that no atten- 
tion w^hatever is paid to drawing room figures on the valve or its 
setting. One reason for this is the possible, disturbing, influence 
of the governor. Prof. Sweet prefers an increase of compres- 
sion to an increase of lead at the head end of the cylinder ; the 
compression will show more on the cards, but this is held to be 
of no practical account as the object is to obtain a smooth run- 
ning engine rather than good looking cards. 

In the smooth running engine for which the diagram was 
drawn, the distribution was practically equaUzed for the two 
ends of the cylinder. So far as this one example goes it 
would therefore seem that uniformity of distribution is favorable 
to smoothness of runninsf. 



STURTEVANT, BALL AND WESTINGHOUSE ENGINE. 169 

Sturtevant Engine, 8x12. 

Fig. 75. The point P of the pendulum arm lies outside of 
the crank radius OC, but Pand ^'are both on same side of shaft 
center 0. The locus LL, struck from /^ as a center, is convex 
to center of its shaft, but it is so located that the lead increases 
with the throw, and the preadmission takes but a small part of 
the stroke. The cut-off varies from \ to I. The piston-valve 
has a diameter of ^" and is of the positive indirect type. The 
steam passages in the cyhnder have a cross-section y^" X ^" . 
The speed of this engine is 250 revolutions per minute. 

Ball Engine. — Recent Type, 12x12. 

Fig. 76. This is the diagram of a more recent type of Ball 
engine than that discussed in Fig. 71. Here the point P is out- 
side of crank direction OC 3.nd Pand C are on opposite sides of 
shaft center 0. The locus LL is concave to (9, but so placed as 
to permit but little change in the lead as the expansion varies ; 
the preadmission also occupies but little of the piston stroke. 
The cut-off ranges from o to ^. In this engine, as in the South- 
wark, a pin takes the place of the eccentric sheave, thus affecting 
a reduction of the work of friction and removing the danger of 
heating. The valve is of the flat and balanced variety ; it exhausts 
at the outer edges and takes steam at the inner ones. 

Westinghouse Standard Vertical Engine, qJ^xq. 

Fig. J J. In this diagram the left-cylinder-crank C is at the 
left and the right one C" at the right, both being placed on the 
vertical line of the engine (not of the diagram). The relative 
positions of crank and eccentric are as they would appear if 
viewed from the right-crank end of shaft. The terms right and 
left cylinder as used here and as used by the manufacturers sup- 
poses the engine viewed from the throttle side ; then it is the 

< I ^ > port that admits steam to the < ,^r, > cylinder. The 



I/O WESTINGHOUSE STANDARD ENGINE. 

center line of each cylinder passes the center of shaft at a 
distance equal to half the length of crank OC". In the diagram 
this distance and crank are drawn to a reduced scale. 

The pivot P of swinging eccentric is tV from the radial line 
of the left crank and has a negative abscissa of 8^''. The 
radius of locus LL is PE^ = qtV. At maximum cut-off the 
eccentricity is i^y^" and the radius PME^ is tangent to the arc M 
described with OM =-. ItV as a radius; at minimum cut-off the 
eccentricity is i/t'' := OE^ and the radial line PmEo is tangent to 
the arc m^ which is struck off with radius Om = i^". The valve- 
circles are found as in Fig. 60, p. 137. The lead v, for all grades 
of expansion, was measured on that dead-center line of crank 
which corresponds to uppermost position of piston and increases 
with the cut-off The engine is composed of two single^ 
acting, vertical cylinders which receive the live steam in the top 
end of the cylinder. As each cylinder has its own crank and 
connecting-rod this arrangement will cause the angularity of 
the rod to have precisely the same influence on the steam distri- 
bution of each cylinder. This is shown by the values tabulated 
on the diagram. The cut-offs evidently range from o to ^. 

The table inscribed on the diagram and valve-circles (in full 
lines) do not take into account the influence of the length of the 
eccentric rod. The vectors of the valve-circles drawn from give 
the hor. distance of the eccentric center En from the line Y'OY' 
which is perpendicular to the dead center line of eccentric. To 
take account of this angularity of the eccentric-rod, we make the 
following construction. Place the crank upon dead point for 
top of stroke and the eccentric in a corresponding position. The 
center E^, will then be 4j^° in advance of position shown in the 
diagram. Taking this new position of E^ as a center and the 
20'' length of the eccentric as a radius, strike an arc that will 
cut the dead-point line of eccentric ; then revolve this eccentric 
through 180° and use the new position of E2 as a center to 
repeat the former operation. Now bisect the distance between 



WESTINGHOUSE STANDARD ENGINE. 



7 



travel to 



f right 1 
I left i 



of center of motion on 



these two arcs and the point of bisection will be the center of 
motion to which the valve travel is referred and from which as a 
center the arc of reference F' V can be described with the 20'' 
radius. The hor. distance of the eccentric center ^„from this arc 
of reference will be the exact travel of valve. If this exact travel 
is laid off on the corresponding crank positions, laying off valve 

f actual "I , 

t prolonged r''^"'^ 

repeating this process for second crank also, we will get two 
pairs of polar curves instead of the two valve-circles, and the 
intersection of these curves with the proper valve-circles 
will give the exact steam distribution. One pair of these polar 
curves, the one for left cylinder crank has been roughly drawn 
on the diagram for maximum throw. They lie on opposite sides 
of valve-circle OD^ which may be regarded as their average 

value. The \ inner I ^^^ ^^ ^^^ polar curves represents travel 

to the \ rio-ht I °^ ^^ center of motion. The exact steam dis- 
tribution given by carefully drawn polar curves is contained in 
the following tabulation : 



Half Throw 


Right Cylinder. 




£1 


= 0.34 
==0.68 


V-.0.2I 
Zj= 0.08 


^1 = 0.38 
-^3 = 0.16 


)y^=0.0I 
ri'^= 0.00 


Half Throw 


Left Cylinder. 




£1 

^2 


==.0.28 

= 0.67 


£/=0.2I 
£/= 0.06 


^x=-o.35 
-^. = 0.17 


5y/=0.00 
)y/z=: 0.00 



The slight inequalities that exist for the two cylinders is 
wholly due to the influence of the length of the eccentric-rod. 
The diameter of the piston valve is \^^' and the speed is 300 
revolutions per minute. 



1/2 VALVE GEAR. 



Westinghouse Compound Vertical Engine, 14 and 24x14. 

Fig. 78. In this case the stroke of the pistons passes directly 
through the center of shaft. The single, piston, valve controls 
two ports M and N (see little figure to right of diagram 78). 
The passage P leading to high-pressure cylinder is always in 
communication with the hollow of the valve. The port M 
therefore admits live steam from pipe vS to the high-pressure 
cylinder, and port iV, at one edge receives steam from high- 
pressure cylinder and admits it to the low-pressure cylinder, 
and at the other edge it delivers steam from low-pressure cylin- 
der to the exhaust pipe E. The steam laps e at the inner edges 
are ahke for both ports Mand N. The exhaust lap is however 
equal to o. The value of e' given in table was obtained by 
multiplying actual lap e by }i (= :j^3, the ratio of the arms of the 
bent lever driving the valve) to reduce it and the stroke of the 
valve to the end of the eccentric-rod. This is a departure from 
our usual practice of assuming that the actual valve is driven 
directly by the eccentric, but in this case it was simpler to change 
only one dimension than to change several. 

The pivot Pdoes not lie on either of the two cranks, but it is 
near to the low-pressure crank and on the same side of center 0. 
The abscissa and ordinate of Pare 5.42'' and 1.5'' respectively, 
its total, radial distance from being S^"- The locus LL 
struck from this center has a radius of 6^/' and is concave to 0. 
The table and data inscribed on diagram state distribution. In 
this Compound, as in the Standard, the steam distribution is the 
same for both single-acting cylinders when the eccentric- rod is 
taken as infinitely long. In this case the rod is finite and 49'' 
long and its influence on the distribution can be ascertained in 
the now well-known manner by drawing the arc of reference 
Y'Y', It is evident from the figure and from what has been said 
about other diagrams, that the valve-circles give the valve-travel 
with sufficient accuracv. 



Fig. 79. 



( 




C B = I' 
C A = lO' 
B E = 6 " 
E F = 8" 

F X -= 8i" 
C X = 5H' 

E L ^ 9i 



C H = 3" 
C 0'= 45f 

H I = jsr 

I J = 12f 
J K = 6" 

L M = 38f' 

M N = nil' 



0'R = 2r 

R P = 3= RQ* 

C R = 52" 



OUTSIDE LAP = O. 

INSIDE LAP - 0.04 

I 

LENGTH OF CONN. 
RADIUS OF CURVEC 



PORTER-ALLEN VALVE GEAR. 1/3 

LINK-MOTIONS. 

In the second group of single-valve gears (see p. 153), the 
eccentrics themselves are invariable, the variation in the valve 
motion being affected by the mechanism between the eccentrics 
and the valves. All the commom link motions belong to this 
group; we will here describe and examine but one of them, 
the Porter-Allen, (or * Fink '), the only one which in this country 
has been extensively used on stationary, high-speed engines.* 

The Porter- Allen is that special case of the Gooch link motion 
which is obtained by supposing the centers of the two eccentrics 
to coincide, thus forming only one eccentric, the two eccentric 
rods and link then constituting one rigid piece. The best pro- 
portion for this linkage are shown in Figs. 79-82, and were taken 
from blue-prints furnished by the South wark Foundry & Machine 
Co., of Philadelphia. 

Porter-Allen Valve- Gear, 111^x20. 

Figs. 79, 80 and 81 give a side view of this engine, a section 
through its rocker arm and a skeleton showint.; the relative posi- 
tion of the three valves when crank is on its forward center. 
This view and the skeleton of the mechanism given in Fig. 82 are 
sufficiently full and complete to render any detailed description 
unnecessary. We will only ask the reader to note that eccentric 
strap and link constitute one rigid piece, that for a particular 
grade of expansion we assume that the governor occupies an 
invariable position vertically, thus making point of suspension / 

* " Prof. Zeuner's views of the unsuitableness of the ' Fink ' link as a 
reversing -gear are correct, and for the reasons he gives. But the use of the 
half-link in the Porter-Allen engines (it is in connection with these engines 
that the Fink motion was first brought into extensive use) shows that when 
compression is judiciously applied, then the cut-off points may be made 
practically 'symmetrical ' up to the half -stroke at the expense of a variable 
lead, the variation of which is in opposite directions for the two strokes. 
Before cushioning was used in these engines the variability of the lead 
made it almost impossible to obtain quiet running under varying conditions." 
12 




C B - 1 " 


C H = 3 


C A - lO' 


C O'- 45? 


B E - 6" 


H 1 - I5j " 


E F = 8" 


1 J - 121 


F X - Si" 


J K - 6" 


C X - 5i^" 


L M - 38| 


E L ~ 9j' 


M N - llH" 



RELATIVE POSITION OF VALVES WITH ENGINE ON FORWARD CENTER. 

LINK BLOCK, (FULL OEAR) 6" FROM TRUNN 
NECATIVE LEAD = sV" V//////////A my/Z/y/Z/A STEAM VALVES. 



C R - 52" 

OUTSIDE LAP- 0.7p 

INSIDE LAP ~ 0.04 

LENGTH OF CONN. ROD - 60" 

RADIUS OF CURVED SLOT - 45H" 




FORWARD CENTER 



EXHAUST VALVE 



174 PORTER-ALLEN VALVE GEAR. 

of the hanger IH a fixed stationary point and finally that even 
then the center G of block does not maintain an invariable dis- 
tance from the trunnion E, but slips a little in its slot. 

The view of the relative position of the valves is only a skele- 
ton and does not represent the multiple port-openings ; each of 
the three valves admits steam at four different places thus offering 
a generous passageway for both the live and exhaust steam. 
Before taking leave of these general views we desire to call 
special attention to the rocker arms ORPQ which communicate 
motion to the rods and stems of the two steam valves. From 
the arrangement we see that each driving arm, 7?/* and RQ' , has a 
period of almost complete rest and two periods of rapid motion. 
The rapid motion period effects a rapid opening of the port to its 
full width and then a rapid closing, the period of comparative 
rest occuring when the port is closed by valve. This arrange- 
ment of bent levers does excellent service but modifies greatly 
the " harmonic motion" of the driving point of the link and re- 
quires other diagrams than the valve-circle for the exact represen- 
tation of the valve-travel. We will return to this matter later on. 

We will first show that the travel of a point on the link EGL 
is approximately like the " harmonic motion " communicated by 
an eccentric that has an infinitely long eccentric-rod. Let A 
represent the distance from center of eccentric B to center of 
trunnion E, the latter joining sustaining arm FE and EGL 
on center line of slot of link ; let p represent the eccentricity CB, 
u the distance of point G above dead center line of eccentric, co 
the crank or (or eccentric) angle and ^ the distance of eccentric 
center B from perpendicular through C to eccentric's dead-point 
line (which is approximately along CEO\ the crank and eccentric 
differing in position by a small angle of 2° 12! that is equal to 
angle included between their respective dead-point lines). Then, 

for the case in which — > 4, we will show that approximately : 

^ =z p cos CO -f —p sin CO (90) 



PORTER-ALLEN VALVE GEAR. 1 75 

To prove this, we consider the total movement of Hnk to be 

composed of two parts, of a horizontal motion in which the link 

moves parallel to itself, the character and extent of this motion 

depending on the horizontal throw of the eccentric, and of a 

rocking motion about the trunnion E (or vibrating fulcrum) which 

is due to the vertical throw of the eccentric. Here the term 

p cos (i) is the horizontal throw common to all points of the link 

u 
and the term -p sin w is the vertical throw ^o sin <^ X the ratio 

11 

-7 of the arms of the bent lever constituting the link and rotating 

about the fulcrum or trunnion E. The formula for ^ thus estab- 
lished is the polar equation of the valve-circle and the coordi- 

11 
nates p and —p of the vertex of its diameter show that this ver- 
tex lies on a rectilinear locus perpendicular to the dead-point line 
of the eccentric and at a distance p from the center of the shaft. 
Such a valve-circle is drawn in Fig. 83 for full gear {i.e., for block 
G about 6' from trunnion), and the corresponding equivalent, 
(virtual or resultant) eccentric can be found in the usual way, 
Fig. 60, from the diameter of this valve-circle.* 

To show that the valve-circle thus found represents with 
considerable accuracy the travel of a point on the link, we 

* As the horizontal component motion due to the rocking, is motion rela- 
tive to trunnion E, we can also combine the parallel and rocking compo- 
nents by means of the parallelogram of velocities. We need only regard 
the rocking component of the motion as directly and independently pro- 
duced by a separate eccentric, one that would follow the actual eccentric by 

u 
90° and have a length —p. As these two eccentrics are both at right angles 

A 
to, and proportional to, the velocities they impart we can construct the 
parallelogram of velocities on the eccentrics as sides and the diagonal will 
give the length and location of a resultant or virtual eccentric capable of 
imparting directly the whole horizontal motion to the point of the link. 
The parallelogram and triangle of eccentricities constructed in Fig. 83 and 
Fig. 84, respectively, are illustrations of this method. The valve-circle can 
then be found from the virtual eccentric as in Fig. 60. 



l!76 PORTER-ALLEN VALVE GEAR. 

have in Fig. 83 compared such a valve-circle with a polar 
curve OMN or OM'N' representing very exactly the travel 
of the rocker-pin 0' . (Figs. 79 and 82.) To simplify matters 
we have assumed pin 0' to travel on a straight line that 
bisected the rise of its arc-path and passed through center C 
(Fig. 82) of shaft. In Fig. 83 we have called this straight 
line CO' (of Fig. 82) the dead center line of eccentric and con- 
structed the valve-circle according to formula 90 found above. 
The dotted, polar, curves 6>J/iVand OM'N' (Fig. 83) were found 
by laying off on the crank positions the exact distance of pin 0' 
from its center of motion, supposing 0' to travel on aforesaid 
dead point line CO' . We see that these curves agree very well 
with the valve-circles and that the latter may unhesitatingly be 
used to represent the travel of lower rocker-pin O parallel to 
dead point line G C Inspection of the setting of the upper 
rocker-pins Q' and P shows at once that the horizontal compo- 
nents of their motion could not be represented by valve-circles, 
and that to represent the valve travel accurately we must con- 
struct exact polar curves for both the front and the back valve. 

A fairer comparison of the actual travel of a point on the link 
with that given by the formula or valve-circle is furnished by 
Fig. 84, which represents by the dotted, polar curves the actual 
travel, from its center of motion, of the point 7^ (Fig. Z2) driving 
the exhaust valve. It is evident that the average direction of 
the point M of the exhaust rocker NM is parallel to the average 
direction of the path of the driving point L on link. We may 
therefore assume that the actual travel of M is almost axactly hke 
that of L. Now the travel of Tis 0.39 of that of M\ it is rep- 
resented by the dotted, polar curves of Fig. 84, which were 
obtained by directly measuring the distances of point T from its 
center of motion and then laying off the distances on the corre- 
sponding positions of the crank. The valve-circle representing, 
approximately, the travel of point L is drawn on diagram on 
diameter ODJ and reducing it 0.39 we get the valve-circles 
OD^,' representing, approximately, the travel of driving point T. 




FRONT VALVE. 



BACK VALVE. 



iVI 



A 



ERNOR TO HIGHEST 
HEN LET GOVER- 
LVE WILL BE 
MORE. 




16 9 





V ** 


^^ 1 


Vjv 


V-^ 


trt 


1 


6i 


-o 


' 


1 
1 

1 


» 


1 

ll<i 





LD BE OPEN I" 

« « 3'/ 



SKELETON OF PORTER-ALLEN VALVE GEAR. 17/ 

Though the deviation of the two sets of curves is now marked, 
nevertheless, the deviations occur where they will but slightly 
influence the beginning of compression or release, which are 
found as of old at the intersections of polar curves (or circles) 
with the lap circles. 

We have already called attention to the fact that the setting of 
the driving arms PR and Q'R causes them to communicate to 
the valves a horizontal motion that is decidedly different from 
the " harmonic motion " assumed for eccentrics. If we wish to 
represent the valve's motion accurately by polar curves whose 
vectors are the crank positions, we must determine accurately 
the distance of the valve from its center of motion for a complete 
series of crank positions. 

Inspection of the skeleton in Fig. 82 shows that it is composed 
of two quadric chains CBEF and HIRO' connected by the 
straight link 6^ F (which in the actual mechanism is represented 
by the link-block G). For the assumed point of suspension / 
the point G of second quadric chain describes the path G, 1,3,6, 
while the point 0' of lower rocker arm describes the correspond- 
ing path 0' , I, 3, 6. In the first quadric chain CBEF the point 
L occupies points i, 8, 13, 23, etc., of its path, while eccentric 
center B occupies points i, 8, 13, 23, etc., of its circular path. 
The center V o\ the slot GE\?> the point on which rod 6^ F turns 
and this point is on the line BE of the first quadric chain; the 
piece BE in the skeleton is supposed to be extended so as to sup- 
port this point V. The path V is given and the numbers cor- 
respond to the crank or eccentric positions. We can easily find 
the position of either valve for any crank position if we know the 
corresponding position of 0' . To get 0' for any eccentric (or 
crank) position, we suppose V placed at that point of its path 
which has the same number as the crank position under consid- 
eration ; with this point Fas a center strike off an arc cutting the 
path of G : then with this intersection of the arc and the path as 
a center and the steam rod GO' as a radius strike off another 
arc ; it will cut the circular path of 0' at the point 0' desired. 



SKELETON OF PORTER-ALLEN VALVE GEAR. 

115x20' 




FRONT VALVE. 



To Set Back Valve. 

SET ENGINE ON BACK CENTER ; RAISE GOVERNOR TO HIGHEST 
POSITION; SET VALVE WITH i" LEAD; THEN LET GOVER- 
NOR DOWN-IF ALL IS RIGHT THE VALVE WILL BE 
MOVED TO OPEN ITS PORT ji" MORE. 




EDGE OF CRANK DISK.) 



To Set Forward Valve. 

SET ENGINE ON FORVBARD CENTER, RAISE GOVERNOR TO HIGHEST POSITION; 
VALVE WITH A' LEAD; THEN LET GOVERNOR DOWN-IF A 
VALVE WILL BE MOVED gi' 1" DIRECTION OF 



178 CENTER OF MOTION OF VALVE. 

The position of each valve can be represented by its point Q 
or 5 and is estimated from a center of motion that is invariable 
in position for all grades of expansion, i.e., for all positions of 
the block G in its slot. To find this center of motion we must 
first find two peculiar eccentric positions near the dead-point 
line of the eccentric. To get these positions we erect at C 
(Fig. 82) the perpendicular CF-, = EF to the dead-point line of 
eccentric. Then with Ft_ as a center and i^ (7 as a radius describe 
an arc tangent to this dead-point line; the chords C\ and (^13 of 
this arc will be the peculiar eccentric positions needed. Now 
find two positions of both Q and 5 corresponding to these two 
peculiar eccentric locations and then find a point half way be- 
tween them ; this point will be the desired center of motion from 
which the travel is to be estimated, and will be found to be the 
same whatever position is assumed for the point G in its slot 
(that is, it will be the same whatever fixed position is chosen for 
the point of suspension / of the hanger IH). The center of 
motion can also be found, and more simply still, in the follow- 
ing way. The steam rod is tV shorter than radius of link slot ; 
near the two intersections of path of center F(Fig. 82) with arc 
of 0' lay off -i^" from arc 0' towards path of F, in the direction 
of middle radius of the link slot, till a place is found (on each 
half of V) where arc 0' and path of Fare just tV apart. When 
center of steam rod occupies either of these two positions of 0' 
the block may be moved from one end to the other of link slot 
without moving rocker pin 0' or either of the two steam valves. 
Hence these two points on arc 0' will give exactly the two 
valve positions that are equidistant from the center of motion ; 
in like manner the two points on path Fthus found will give the 
corresponding eccentric positions and these will be found to be 
near the pecuHar ones described above. The distance of each 
valve from its center of motion is now laid off on correspond- 
ing crank positions and the polar curves shown in Figs. 86 and 
Zj for full gear (6^ is then 6" from trunnion E^ obtained, also the 
polar curves shown in Figs. 89 and 90 for half-gear (block G is i" 
from trunnion E). To ascertain the lap from the data given in 



Figs. 89, 90, 91. 

MS Foa Both Steam Valves 



32 



?ANKS 



LINE' 






= 6 



Half Gea! 



FRONT valve: 




LINE OF CRANK 



BACK VALVE 



C.LINCOF CRANK 




LAP or BACK valv/e: 



BACKy 



MOTION CURVES. 179 

Figs, 79-82, we must find positions of point ^ c r when engine 

is on \ back 1 ^^"^^^ ^"^ block G at trunnion £; as the edge 

of the j 11. > steam port is then tV < ri^yht I ^^ ^^^^ °^ 

{ bTc"k } ^^^^"^ ^^^^^' ^'^ ^^>^ ^^^'^'' ^° ^^^ { right } ^^ { 5 f 
to get its position when valve edge is exactly at port edge. 

Now place ^ ^^ [• at its own center of motion and measure its 

distance from its former position when valve and port edges coin- 
cided ; the distance wifl be the steam lap desired. Laying off the 
same steam lap = ft for both front and back valve, and measuring 
in each case the distance RS (Figs. 86-90) and dividing it by the 
stroke = (2 X CO), we get the cut-off effected at each end and at 
each grade of expansion. The shaded areas give in all cases the 
linear port openings and should be multiplied by 4 for the total 
openings. The numerical values of the distribution are inscribed 
on the diagrams and show that it is a practically perfect valve 
gear in this respect. 

The actual valve travel found above can also be laid off as 
ordinates, on the piston stroke as a base, and then we get another 
series of valve diagrams of an oval shape, that are known as 
"motion" curves. The exact polar and oval diagrams give of 
course identical results. Figs. 85, 88 and 91 respectively repre- 
sent the travel of the exhaust valve, that of the front and back 
steam valves for full gear and that of the front and back 
steam valves for half gear. In Fig. 85 cushioning in crank end 
is measured by aX ^- OX, release in head end by dX -7- OX; 
similarly cushioning to head end is given by dn -^ OX and re- 
lease in crank end by cj/i ^- OX. In Figs. 88 and 91, ad -^- OX 
measures the cut-off in crank end and Nc ^- OX the cut-off in 



Figs, 83, 8,, 85. 

-Diagrams of Rocker-Hin O' 

FULU GEAR ^ ,^Jl^*-?'°'^"'°°< 



Figs. 86, 87, 88. '^''' 

-Exact Fouar and Oval Diagrams fob Both Steam Valves 

Half Geafi 



FaoNT \/alve: 




l80 DOUBLE- VALVE GEARS. 

head end. The polar diagrams for the steam valve best show 
the variation of lead. The speed of this size of Porter-Allen 
engine is 230 revolutions per minute. 



Double- Valve Gears.* 

The characteristic of a double-valve gear is that while the first 
of its two valves slides on a fixed seat, the second sHdes on the 
first, the motion of the second valve relatively to the first being 
affected by the absolute motion possessed by each of the two 
valves. 

In single-valve gears with variable elements, and in link- 
motions, we simplify our problem by finding the equivalent or 
virtual eccentric and then proceed as if the valve were directly 
moved, to and fro upon its fixed seat, by this virtual eccentric. 
In double-valve gears a like simplification is introduced, in fact 
it is the finding of this virtual eccentric which usually constitutes 



* This term, double-valve, has occasionally been applied to a valve con- 
sisting of two rigidly connecting halves and also to two separate and 
independent valves each sliding on a fixed seat. The gear here called 
double-valve gear has sometimes been defined as gear with plain slide 
valve and independent cut-off; it has also been described as composed of 
a plain slide valve and a riding valve. But these are long expressions and 
are also not free from ambiguity. 

We wish to distinguish between gears requiring considering of only 
absolute motion and those in which relative motion must be considered. 
Gears possessing but a single valve evidently belong to the former class 
and the term, single valve, is sufficiently suggestive of absolute motion. As 
relative motion implies at least two pieces we feel justified in using the 
term double-valve to suggest this motion. A further justification is sup- 
plied by the fact that the principal functions, of opening or closing a steam 
passage, must be doubly performed (though not simultaneously) when one 
valve slides on the back of another. We must confess, however, that we 
are not satisfied with the terms we have chosen and hope that something 
else as simple but more exact may be found. 



PARALLELOGRAM OF ECCENTRICITIES. 



I«I 



the main part of the problem. Such an eccentric drives the ex- 
pansion valve, over a perfectly stationary main valve, with an 
absolute motion that is the same as the motion of the expansion 
valve relatively to the main valve when both these valves ha\'e 
their actual motion. In this way the new and complex problem 
of the relative motion of two valves, is reduced to the old and 
simple problem of a single-valve sliding on a fixed seat and 
driven by one eccentric. 

Now as to the method of procedure. When the main and 
expansion eccentrics are both given, the virtual eccentric can 
easily be found by means of a proposition for which Dr. Zeuner 
suggests the nam>e of " Parallelogram of Eccentricities." 




Fig. 92. 



It may be stated as follows : The main, virtual and expansion 
eccentrics constitute^ respectively, the tzvo adjacent sides ajid diag- 
onal of a parallelogram. 

In Fig. 92, OE„i, OEy and OEe represent these eccentrics and 
OEynEgE^ the parallelogram which they form. To avoid confu- 
sion hereafter it should be specially noted that the expansion 
eccentric is always the diagonal of this parallelogram, provided 
the virtual eccentric OE^ effects the m.otion of the expansion 
valve relatively to main valve. But when it is a question of 



1 82 DESIGNING DOUBLE- VALVE GEAR. 

relative motion of main valve to expansion valve the virtual 
eccentric must be OEv\ just opposite to OE.v, and OE^ must 
then be the diagonal of parallelogram OEe OE^, OEJ. In all 
our work we shall take the expansion eccentric as the diagonal of 
the parallelogram. 

In existing ysXyg^ gears the main and expansion eccentrics are 
completely known and the examination of the steam distribution 
effected by their valves is easily made by means of the main and 
virtual valve-circles, which are deduced in the usual way from 
the known main and virtual eccentrics. 

In designing valve gears, however, the expansion eccentric is 
the unknown quantity whose variations must satisfy the pre- 
scribed conditions as to range of expansion, time of admission, 
rapidity of cut-off, etc. To find it we must first find the main 
and virtual eccentrics and with these two as adjacent sides con- 
struct a parallelogram whose diagonal will be the desired expan- 
sion eccentric. 

The main eccentric can be found in the usual way from its 
valve-circle and this in turn from the given conditions as to lead^ 
port-opening, maximum cut-off, release or compression. The 
virtual eccentric is found in a similar way, its circle being first 
obtained from such given conditions as, type of expansion valve, 
cut-off and time of reopening. 

Having thus indicated the principal steps in designing a 
double-valve gear, we will return to the beginning and show 
that, when there are two valves and both are in motion, one rid- 
ing on the other, then a fixed circle can be found which will cut 
from the crank, or its prolongation, chords whose lengths repre- 
sent the distances between a point on the expansion valve and a 
corresponding point on the main valve, the correspondence being 
simply that the two points are together when the expansion 
valve occupies its middle position on its seat on the main valve. 
The distance between two such points at any crank position is 
evidently equal to the (algebraic) difference of their distances 
from any common, fixed, reference line. Neglecting the angu- 



VIRTUAL VALVE-CIRCLE. 



183 



larity of the eccentric-rods, this difference is equal to the differ- 
ence in the distances EgSe and E^nSm, Fig. 93, of the correspond- 
ing eccentric centers Ee and E„i from their common reference 
line OP, which is perpendicular to the dead-point line NOZ 
of the eccentrics. Our problem is now reduced to finding a 




*Fig- 93. 

circle that will cut from any crank position OC 2. chord equal 
to the corresponding difference EgSe — EmSm = E^S of the 
distances of the centers E^ and E,,^ from the middle position 
OP of the eccentrics. 

Finding the main valve-circle 0D„^ from its eccentric 0E„, as 
in Fig. 60, p. 137, we get the chord OE^^^ = 0E„, = E„^S^ as 
the distance of E„t from OP. The distance EeS^ of Eg from OP 
can be found in the same way by constructing the expansion 
valve-circle ODg. To do this we measure angle (7g, from the 
rz£^/U portion the eccentric's dead-point line OZ, in direction 



*The inner angle c^, extends from (9Z around to OB^,. The outer angle a^ 
extends from OC around to ODv. 



184 VIRTUAL VALVE-CIRCLE. 

opposite to the rotation, and lay off this angle a^ from the crank 
in the direction of rotation ; ODg = OEg will be the diameter of 
the expansion valve-circle and OFJ = OFe = EeSg. Then be- 
cause D„^S" is parallel to crank position OC, we have DmS" = 
FJFJ = FgF.n = EeS. As D^FJ is perpendicular to OC, the 
angle D^^S" De is a right angle. But the hypothenuse DgD^ is 
constant in position and magnitude for all crank positions ; the 
Iccus of S" is therefore a circle. If through D„i we draw a par- 
allel to the crank position OC 2X. any instant, this circle will cut 
from this parallel a distance D,nS" which is equal to the value 
of EgS at this instant. By moving this circle parallel to itself so 
that the point D^ of this circle D^S' Dg shall shift its position to 
shaft center (9, the circle will take up the new position OD^FJ 
and cut from the crank itself (or its prolongation) the distance 
OFJ = D^S" = EgS = EgSg — E^nSm- In its new position 
it may be called the virtual valve-circle for it possesses all the 
properties of a valve-circle relatively to its eccentric. For in- 
stance, the virtual eccentric OE^ is found from the virtual valve- 
circle OD^ by laying of the angle a^, in the same way as the 

angles a„i and Og. Moreover, when the \ orolonp-ed 1 ^^^"^ ^^^'^ 



the virtual valve-circle, the virtual eccentric will be to the 
of its middle position OP. As regards the periods 



( right ) 
I left f 



during which the riding-valve keeps the ports in the main 
valve opened or closed, the virtual valve-circle, like the two 
others, is subject to the conditions of valve-type, etc., laid down 
in Table XIV. The valve driven by the virtual eccentric is of 
course of the same type as the expansion valve, the relative 
motion of the latter corresponding exactly to the absolute motion 
of the former. 

Before applying these results to the common high-speed 
engines, there is still another proposition enunciated by Dr. 
Zeuner which is of the greatest service in the solution of valve- 



VARIATION OF EXPANSION. 



185 



gear problems. But before stating what it is, we will point out 
that with double-valves the expansion can be varied : 

I. By varying the angle of advance of the expansion eccentric. 
II. By varying the throw of the expansion eccentric. 

III. By varying both angle and throw of the expansion eccentric. 

IV, By varying the lap of the expansion eccentric. 

In the last case neither throw or angle of advance of expan- 
sion eccentric is varied. The most general case is therefore III ; 




^— X- 



Fig. 94. 



it is represented in Fig. 94, where the center E^ traverses any 
arbitrary locus ac. Its reference line is taken to be the crank 
at left dead point, but any other crank position might have been 
chosen and the locus ac would occupy a correspondingly different 
position in the plane of the paper. It is simplest to consider the 
locus ac as rigidly attached to the crank, moving with it, and 
preserving always the same relative position towards it. 

Let OX and (9 F be coordinate axes rigidly attached to the 
crank and moving with it, then will OF and FEg represent the 



1 86 



LOCI OF CENTERS AND VERTICES. 



coordinates of any point Eg of the locus ac. Prolong main 
eccentric EmO to 0' , making 00' = OE^^, and draw a parallel 
set of coordinate axes 0'X\ 0' Y' , also rigidly attached to crank. 
Then the proposition of Dr. Zeuner, last referred to, consists in 
the statement that, The locus a'c' of the center Ey of the virtual 
ecce7ttric OEv is a curve exactly equal and parallel to the locus ac 
and the position of 3!c' relatively to the origin O' is exactly like 
that of 2.0, relatively to origin O. 

To prove this it is only necessary to show that the coordinates 
0'F\ F'Ey are respectively equal and parallel to coordinates OF 
and FEg. By construction 00' is equal and parallel to EgE^^ 
which makes OO'E^Ee a parallelogram, hence OEg is equal and 
parallel to O'E^ and this makes the two sets of coordinates 
respectively equal and parallel. Since in designing we start with 
the valve-circles rather than with the eccentrics it will be conve- 
nient to represent this prososition as applied to the diameters of 
the valve-circles. 

Y 




^Mf''^^^\yj 


\ 


"""J^K ^-^"''^^ / 


\ 


J^ N ^y/ 


\ 




\ 




\ 




\ 


y' ~ . 




A 





i' 



Fig. 95- 



In Fig. 95 the curve ac is the locus of the vertex D^ of the 
valve-circle diameter ODg. This curve ac is equal to the locus of 



LOCUS OF CENTER OF ECCENTRIC. 1 8/ 

Eg and is similarly located. The curve dc' is the locus of the 
vertex D^ of the diameter OD^, of the virtual valve-circle, and is 
equal and similarly situated to the locus of E^. The locus a'c' 
has the same position relatively to the origin 0' that ac has rela- 
tively to the origin 0. These valve-circle-loci, unlike those in 
Fig. 94, have the great advantage of preserving their places in 
the plane of the paper during the rotation of the crank. 

The subscripts e, in, v employed in these diagrams relate, re- 
spectively, to expansion, main and virtual eccentrics and their 
valve-circles. The symbols^o, r^ and ^o respectively refer to the 
lap of the expansion valve, the eccentricity and angle of advance 
of the expansion eccentric. When the subscripts i and 2 are 
annexed to the subscript o the values at minimum and maximum 
grade of expansion are meant. The other symbols have the 
meaning already assigned them in the discussion of single-valve 
gears. 

The loci of the center of expansion eccentric vary greatly in 
form according as the valve-gear elements, angle of advance, 
throw and lap are varied. In Figs. lOO and loi, representing 
the expansion valve diagrams of the Cummer engine, the locus 
of ^^ is circle concentric to shaft center and has r^ as a radius. 
In Fig. io6, the expansion valve diagrams for a Meyer valve 
gear, this locus reduces to a point, for the setting and throw of 
the expansion eccentric does not vary with the grades of expan- 
sion but only with the lap. The inscription, " locus of D^,' on 
this diagram is to be understood only as a geometrical construc- 
tion that is helpful in solving the problem. In Fig. 109, the ex- 
pansion valve diagram of the Rigg engine, the locus ^^ is 0EejEe2, 
a straight line passing through 0. In Fig. 1 12, the locus of ^^ is 
a circle that is not concentric to 0. In Figs. 115, 116 the locus 
Eg is a straight line that does not pass through and in Figs. 
118, 119 locus of ^^ is approximately a circle whose circumfer- 
ence passes through shaft-center 0. 

In designing a double-valve gear, the steps are : [a) to find the 
main eccentric suitable for release, compression, and beginning 



I 88 EXACT VALUES OF VALVE TRAVEL. 

of admission, {b) to find the virtual valve-circle that will give the 
desired cut-ofF, reopening, etc., for the assumed type of expansion 
valve, {c) to find the corresponding virtual eccentric, and {d) to 
find the desired expansion eccentric by combining the main and 
virtual eccentrics. 

When very great accuracy in the determination of valve travel 
is desired, we must substitute for the virtual valve circle a pair 
of exactly drawn curves whose vectors represent at once the 
crank's position and the valve's travel from its center of motion 
(see definition of this center, p. 151). To construct these polar 
curves we must therefore find this valve travel. Accordingly we 
should, in Fig. 93, replace the middle position or reference line 
P OP of the main eccentric by a reference arc struck from the 
center of motion with eccentric rod as radius. The distance of 
OE^ from this arc will be E^SJ. In like manner we must 
replace the reference line POP oi the expansion eccentric by the 
reference arc (of the valve) struck from the corresponding center 
of motion and the eccentric rod as radius. The distance of Et 
from this arc will be EgSe and the true relative position or travel 
of the expansion valve will be EeSJ — E^^SJ and this difference 
laid off on the corresponding crank position will give one point 
on the desired polar curve. 

The polar curve obtained by laying off on the crank the dis- 
tance E^S'^ (of center E^ of the virtual eccentric from the 
common arc of reference) is in general not a closer approxima- 
tion than the valve-circle itself. 

When there is a reversing lever between one valve and its 
eccentric, the actual distance of each from the arc of reference 

•1, , , , • 1 11 f difference) 

must still be taken, remembermg however to take the j \ 

of EeSJ and E„^SJ when the valves are on the | ^ppo^^ite^sides ) 
of their center of motion. 



Figs. 96, 97, 98. 

B UCKEYE CNGINC . 

Expansion Varied by Angle of Advance. 



Main Valve Diagram, 



ansion 



Positive Indirect Vaiect Valve 




Expansion Valve Diagram. Negative Direct V^RECt Valve 



Minimum Cut-off =0 




UT-OFF=0.S 



Expansion Valve Diagram. Negative Direct V/rect Valve, 
Maximum Cut-off = That of Main Valve. 




Reversing Arms between Eccentric and Val 



EXAMPLES OF DOUBLE- VALVE DIAGRAMS. 1 89 

In the great majority of cases the valve-circles will be suf- 
ficiently accurate ; in the few remaining cases it will rarely be 
necessary to construct the whole polar curve ; a few points of 
the curve (for the crank positions of the most practical interest) 
will usually suffice. 



Examples of Double- Valve Diagrams, Figs. 99-119. 

In these mechanisms the expansion valve moves on the main 
valve and controls the ports or steam passages in the latter. Since 
both valves move their relative motion depends upon the separate, 
absolute, motions and is one of some complexity, but it can be 
simplified by considering the main valve stationary and supposing 
the expansion valve moved by a new special eccentric that im- 
parts to it an absolute motion equal to the relative motion which 
it possesses when both valves move under the influence of their 
own, actual, eccentrics. This eccentric has been called the vir- 
tual eccentric and for a particular grade of expansion has all 
the simplicity and properties of the ordinary fixed eccentric. 
Its valve-circle is found from the virtual eccentric in the usual 
way and gives the same sort of information concerning the 
crank positions at which the port in the main valve opens and 
closes. 

In this set of diagrams the port openings have not been indi- 
cated by sectional areas. With negative valves the port open- 
ing is measured inward; from negative lap-circle to valve-circle, 
this intercept lying wholly or partly outside of the valve-circle. 

When there is a reversing arm between an eccentric and its 
valve, we assume in diagramming the same type of valve but an 
eccentric capable of giving the valve motion directly, without 
such an intermediate lever. At the end of the procedure this is 
allowed for by placing the resultant expansion eccentric 180° 
from that given by diagram. 

13 



BUCKEYE ENGINE CUMMER ENGINE. M EYER VALVE GEA R. R IGG. ENGINE . ST URTEVANT ENGIN E. GUI NOTTE VALVE S EAR. BIL GRAM VALVE GE AR. 

■ _ — ; _ ^ „.„.,„..,„.„,.. Expansion Varied Bv Throw OF Exp.NsioN Expansion Varied bv.Simultaneous Variation _ Expansion Varied by a Mechanism that is _ Expansion Varied by a IVIechanjsm that is 

Expansion Varied BY Angi 




190 BUCKEYE ENGINE. 

As this series of figures (99-119) is intended to illustrate the 
double-valve system of varying the expansion, we have neg- 
lected the exhaust side of the distribution in our main valve 
diagrams. But this may easily be supplied from what has gone 
before. 



Buckeye Engine, 13x24, Figs. 96, 97, 98. 

We have already touched upon the kinematic character of 
these valves in the foot-note, p. 154. 

Generally when there are two valves there are two eccentrics and 
these usually give a resultant or virtual eccentric, but not always. 
There is one exception in the Buckeye Engine. In this case the 
mechanism between expansion eccentric and expansion valve is 
such that the rod of this expansion valve receives a motion 
nearly equal to that of the main valve thus making the relative 
motion of the expansion valve on the main valve dependent 
simply on the motion of the expansion eccentric. The latter 
therefore is at once expansion and virtual eccentric. In this case 
of double valves and double eccentrics we need speak only of the 
expansion eccentric and valve-circle. 

Fig. 96. The main valve in this case covers the port in its 
middle position and cuts off with its inner edges which brings it 
into the Positive Indirect class of valves. The location of the 
corresponding valve-circle OFD^n, Fig. 96, can easily be chosen 
by the help of Fig. 62 and the remarks made on pp. 142, 143. 
The eccentric can now be found as in Fig. 60. It is evident from 
Table XIV that left cylinder port is open for steam admission 
between crank position 3 and 4 and closed during crank's motion 
through positions 4 — i — 2 — 3, closing at 4 and reopening 
at 3. 

Fig. 97. For the sake of economy of construction we will 
take ro = r and describe a circle with as center and r as a 
radius. To find the valve-circle for minimum cut-off, when neg- 
ative lap £0 is given, we erect at //" a perpendicular HDg^ to dead 



BUCKEYE ENGINE. I9I 

center line OH of crank. Then will ODe^ be the diameter of 
the expansion valve-circle for minimum cut-off. Inspection of 
Table XIV and Figure 97 shows that this negative direct valve 
keeps its left port in the main valve closed from crank position 
3 to 4 and opening during 4 — i — 2 — 3, the reopening taking 
place at 4 after main valve has closed the left cylinder port. In 
getting the position of the expansion eccentric it should be 
noticed that in this particular problem the dead center lines of 
eccentric and crank do not coincide differing by about 8^°. After 
the manner of Fig. 60, we find OEt^, to be the position of expan- 
sion eccentric at minimum cut-off, when crank is at left dead 
point. As there is a reversing lever between eccentric and valve, 
we must reverse OEg^ to get position OE' e^ of expansion eccen- 
tric in the actual case. 

Fig. 98. In order that the expansion may have the widest 
possible range the expansion valve at maximum cut-off should 
close its port at the same instant that the main valve closes the 
corresponding cylinder port. To reahze this place the crank at 
position belonging to main-valve's cut-off and then find, accord- 
ing to Fig. 62, the location of the proper valve-circle. As the lap 
€0 does not change the perpendicular IDe^ to the crank position 
C0\ will give ODe^ as the diameter of the valve-circle desired. 
Laying off the proper angles as in Fig. 60, we get OE^^ for the 
position of expansion eccentric at maximum cut-off, when crank 
is at cut-off position and no reversing lever exists between eccen- 
tric and valve. As there is such a reversing lever, we must change 
by 180° the position of OEe^ and thus get OE'^^, the position of 
the actual expansion eccentric when crank is at position of maxi- 
mum cut-off. For this same position CO of the crank the main 
eccentric is at OE,,^ and the expansion eccentric's position cor- 
responding to minimum cut-off is at OE g^. The angles of ad- 
vance :0o foi" the extreme cut-offs are easily obtained provided the 
definition of this angle, given on p. 150, is borne in mind. The 
angle Eg^ OB e^ measures the variation of angular position of 



192 CUMMER ENGINE. 

expansion eccentric for the whole range of cut-off. Table XIV 
again shows that the left port in the main valve is open during the 
crank's motion 4 — i — 2 — 3, and there will therefore be a 
large lead by the expansion valve at the beginning of the for- 
ward stroke. The cut-off given on this set of diagrams are 
approximate only. They can be obtained more accurately from 



L 
R 
this connection see method of finding cut-off in Fig. 63. 



the accompanying piston travel diagram in which — = 6. In 

A. 



Cummer Engine, 11x20, Figs. 99, 100 and ioi. 

The functions of the steam distribution are divided among 
three pairs of valves, three valves for each end of the cylinder. 
There is a so-called main valve which however only regulates 
he beginning of the admission of steam to the cylinder, though 
It also has some influence on the relative velocity with which the 
cut-off valve opens and closes its ports in the main valve,. Next 
there is the cut-off or expansion valve which as its name implies 
regulates only the grade of expansion. Finally there is the 
exhaust valve which regulates the beginning of release and com- 
pression for each end of cylinder making them both constant for 
all grades of expansion. In each of these three pairs of valves 
the two halves are rigidly connected. There are at each end 
three steam ports and fozcr exhaust ports, all valves being of the 
gridiron variety. None of these flat valves are balanced, be- 
cause the valve stroke is small and therefore the work of friction 
so small that there would be little gained by balancing. The 
port openings are large for both admission and exhaust. 

In this engine the expansion is varied by changing the angular 
position of the eccentric, as in case of the Buckeye Engine. 
In the case of the Cummer Engine however, the expansion and 
virtual eccentric are not identical, the latter being a side of the 
parallelogram of eccentricities constructed on main and expan- 



CUMMER ENGINE. 1 93 

sion eccentrics, as in the full lines of Fig. 92. Here the complex 
motion of two valves is reduced to one valve driven by an equiv- 
alent eccentric called the virtual eccentric. When this is known 
the expansion eccentric is known and the problem is solved. 
The principal steps of the solution are : (a) To find the locus 
of the vertex of the diameter of the virtual valve-circle, (d) To 
construct these valve-circles for the extreme cut-offs, {c) To find 
the virtual eccentrics from the diameters of the valve-circles. 

Fig. 99. The construction of the main valve diagram presents 
nothing new. We may therefore omit a description of it. We 
simply call attention to the fact that with this positive, direct, 
main, valve the left cylinder port is open from crank position i 
to position 2 and closed more than half the revolution, namely 
2 — 3 — 4 — I, (see Table XIV). This and the expansion dia- 
grams are drawn to twice the usual scale because of the smallness 
of the elements of the problem. The lead v is only ts" instead 
of tV as given in diagram. The width of port is ^''. 

The exhaust valve's diagram has not been given, though it 
deserves attention because there is a reversing lever between it 
and the eccentric of the main valve. This eccentric drives both 
the exhaust and the main valve, but the latter is driven directly. 
These two valves are both of the positive direct type, but move 
always in exactly opposite directions. If the exhaust valve kept 
its present motion but were driven directly its eccentric would 
have to be 180° from its present position and its valve-circle 
would be drawn 180° from the equal one shown in Fig. 99. But 
this figure, as it stands, may be used for the exhaust valve pro- 
vided we (draw the inside lap and) remember that then the 

-J , , > crank cuts the present valve-circle when exhaust 

valve is to the < • hf c ^^ ^^^ middle position. The lap on the 

exhaust side may easily be ascertained by assuming that cushion- 
ing takes place during the last fifth of the stroke ; the beginning 
of the release can be determined from the inside lap thus found. 



194 CUMMER ENGINE. 

Fig. 100. Given e^ and r^. As the eccentricity r^ of the 
expansion eccentric does not change, the locus of the center of 
the latter will be a circle FLM described about with radius r^. 
The vertex of the diameter of the expansion valve-circle has the 
self-same locus i^Z^J/ about 6^ as a center. Consequently accord- 
ing to principle enunciated in connection with Fig. 95 the locus 
of the vertex D^ of the diameter of the virtual valve-circle must 
be a circumference NQP struck with Vo as a radius and 0' as a 
center. As the negative lap is given, we have only to erect the 
perpendicular HD^t_ to the crank position C OH at minimum 
cut-off to get a second locus of D^j^ and thus get the diameter 
OD^^ of the desired valve-circle. We see that the reopening of the 
expansion valve takes place just a little before the crank reaches 
the dead point. By laying off the angles as in Fig. 60, we find 
OE^^ as position of virtual eccentric for to cut-off and then, 
by means of the parallelogram, OEe^ as the position of the cor- 
responding expansion eccentric when crank is at dead point. 



Fig. 10 1. Given e^, To and the extremes of cut-off In like 
manner we find OEe^z is the position of expansion eccentric at the 
maximum cut-off If we transfer the eccentric position OEg-, in 
Fig. 100, for minimum cut-off and crank at dead point, to the 
present figure, the angle Eg^^ OEg^ will measure the total change 
of angular position of the expansion eccentric for this range of 
cut-off 

Two limits may be placed upon the time of reopening of the 
expansion valve. The first hmit is set by the condition that the 
reopening of its port at either end by the expansion valve must 
surely occur after the main valve has closed the steam port at 
that end ; otherwise there will be two admissions of steam during 
the same stroke, and the benefit of expansion will be destroyed. 
The second limit is prescribed by the condition that, for the 
same cylinder end, the expansion valve must open port as soon 
as, or sooner than, the main valve. The two crank positions at 
which the positive main valve cuts off and admits steam at the 



CUMMER ENGINE. 195 

same port, are always more than i8o° apart. Hence, with a 
negative direct expansion valve, we have only to guard against 
a premature opening and therefore need to consider only the 
first of the two limits mentioned above. 

Let us assume new data, namely, that r^ is given, also the 
extremes of cut-off Sq^ and e^^ and the crank position at which 
main valve cuts off; required a suitable lap Co. 

Now the diameter of a valve-circle lies half way between the 
two vectors that contain its intersections with the lap circle. If 
for minimum cut-off, we bisect the angle between the crank at 
this cut-off and the crank position when main valve closes, and 
prolong this bisector till it cuts the given circular locus, this 
bisector will constitute the limit to OD^^. A perpendicular 
dropped from the intersection B^^ of this bisector with the circu- 
lar locus will give the maximum lap ^o that can be used in this 
gear with a negative direct valve. Such a lap will give the 
largest port openings, but if these are already ample a smaller 
lap ^o niay be chosen which will favor the average cut-off in the 
matter of rapidity of closing. In Figs. lOO and lOi a smaller lap 
than the maximum given above is used. 

Let us again assume a new set of data in which the lap e^, the 
crank positions at the extreme cut-offs of the expansion valve 
and at the closing of the main valve are given, while the radius 
^o of the expansion eccentric is required. 

We know from the nature of the gear that r^ is constant and 
that 0^ is the center of the, as yet unkown, circular locus of 
Z)^. Where the lap circle cuts the prolongations of the crank 
positions corresponding to minimum cut-off and closing of main 
valve, erect perpendiculars to the crank and find their intersec- 
tion D^^. Then 0' D^^ will be the radius of expansion eccentric 
which can be employed with the given lap c^. 

The small laps of the expansion valve and the small throws of 
its eccentric in this case are due to the gridiron type of valve 
used. It is one of the good points of this engine that it thus 



196 MEYER VALVE GEAR. 

cuts down the work of valve friction without decreasing the 
excellence of the admission. 

The cut-offs can be found more exactly from the accompany- 

L 
R 



ing diagram of piston travel, for which — is again 6. (See 



method used in Fig. 63.) 



Meyer Valve Gear. 

In this arrangement the expansion eccentric preserves an 
invariable throw and an invariable position relatively to the 
crank, the expansion being varied by changing the amount of 
the lap of the valve. The mechanism for doing this is shown in 
all the elementary text-books of the steam engine and concern- 
ing it we will simply say, that the expansion valve consists of 
two halves that may be moved apart to any desired extent by 
means of right and left-hand screws located on the rod connect- 
ing these two halves. The separation may be carried so far as 
to change the valve from one having negative lap to one having 
positive lap. As the expansion eccentric does not change at all 
with the grade of expansion, there is in this case no locus of the 
center Eg of the expansion eccentric and then, strictly speaking, 
(in an existing gear), there is no locus of the center D^ of the 
virtual eccentric. The locus of Z^^, inscribed on Fig. 106 is simply 
to be regarded as a geometrical help in finding the angle of 
advance ^o when the eccentricity Vo is known. As the expan- 
sion, like the main, eccentric is fixed as regards its setting and 
throw, there will be only one parallelogram of eccentricities (or 
of diameters), only one virtual eccentric and only one virtual 
valve-circle. 

Before solving any problem in connection with this gear it 
will simplify matters to first demonstrate a couple of auxiliary 
propositions. The first of these relates to the finding of the locus 
D^iox a special case and the second to the equalization of the 
cut-off in the Meyer gear. 



MEYER VALVE GEAR. 



197 



It can be shown that, if the difference e^ — e^ = E is pre- 
scribed or given by the proportions of the expansion valve, the 
vertex D^ of diameter OD^ of virtual valve-circle must lie some- 
where on a straight line SV, Fig. 102 and Fig. 106, parallel to 
the bisector OQ o( the angle (p (= NOL) included between the 
crank positions for maximum and minimum cut-offs, the distance 
of the locus from the bisector being 



SR = -^^ 



2 sm — 



{91: 



Let OL and ON represent the prolongations of the crank posi- 
tions at the extreme cut-offs, OQ the bisector of the angle 



VH 




^ = LON and KL = MN =. PT = E = e, — e, = given dif- 
ference of outside laps. Then 



SP = -^— and SR = SP cos t= — — 
sm (p 2 

^ 2 sm 



198 



MEYER VALVE GEAR. 



The remarks and constructions given on pp. 194, 195 in con- 
nection with the Cummer engine, are also applicable here. 

As equality of cut-offs is conducive to smooth running, Dr. 
Burmester has given a solution of the problem of equalization 
for the Meyer valve, which consists in making the laps unequal 
for the two ends of the valve, the difference between the laps 
being a constant quantity. 




Fig. 103. 

The procedure is as follows : In Fig. 103 the stroke FoF° is 
divided into eight equal parts and through these divisions, with 
the connecting rod as radius, arcs are struck intersecting the 
crank circle at i^, F^, F^, etc., then Z^ F^, Z^ F^, Z^ F^, etc., 
represent ^, ^, ^8, etc., of stroke, or cut-offs, and OF^, OF^, OF^, 
etc., the corresponding crank positions. These cut-offs j^,%, etc., 



MEYER VALVE GEAR. 



199 



are laid off as abscissas in Fig. 104. Now let the valve-circle 
for this case be represented by OQ^ Q^\ it will cut from the 
crank the chords OQo, OQ^, etc.; these represent the travel of 
any point of the valve and are laid off as ordinates T^vi^, T^m^,, 
in Fig. 104. 




Fig. 104. 



Each of these chords also represents the lap of the valve 
effecting the cut-off at the crank position on which the chord 
lies. Hence the ordinates of curve m^m^nir, . . . iriz also repre- 
sent the laps effecting the cut-offs represented by their corres- 
ponding abscissas, the ordinates above the base representing 
positive, and those below negative, laps. In like manner the 
ordinates of curve in'^ni^nf- , . . . nf represent the laps effect- 
ing the cut-offs during the return stroke, It is evident from the 
figure that for the same cut off the laps for the return stroke are 
greater than those for the forward stroke. In this diagram these 
differences can be made to (nearly) disappear by moving the 
lower curve upward through a distance m'^m^; it will then 
occupy the position shown by the dotted curve m^m^. In the 



200 MEYER VALVE GEAR. 

mechanism itself, the corresponding step is to start with a sym- 
metrical valve having the same lap on its two parts, then set the 
right-hand half of the expansion valve, nearer to its left-hand 
half by just this amount m^m^. Approximately equal cut-offs 
will then be effected on each stroke. 

It is evident from the diagram, Fig. 104, that a straight line ee 
will approximately represent the two-lap curves between the cut- 
offs yi and ^. To equaHze the cut-off it is therefore unnecessary 
to move the two halves by two screws of unlike pitch. Both 
screws may be of the same pitch provided the two halves of the 
expansion valve are placed on their seat on the main valve with 
the unequal laps given above. 

The method of finding the unknown elements in this (Meyer's) 
valve gear of course varies with the character of the data. We 
shall assume in all the special cases discussed, that the range of 
cut-off of the expansion valve is given and that the laps, setting 
and throw of the main valve are known. We therefore know the 
crank position at which main valve cuts off and care must be 
taken that the expansion valve does not permit a second admis- 
sion of steam during the same stroke. Although the locus oi Ee 
(and locus of D^ is reduced to a point for this gear, it still remains 
true that the point D^ must lie at the distance r^ from the point 
0'\ this point, we know from Fig. 95, is found by prolonging 
OD^ and making 00' = OD^. The problems and their solu- 
tions are arranged in the following order : 

(i.) Given r^ and ^o; required e^^ and e^^. 

(2.) Given To and e^^ ; required ^o and e^^. 

(3.) Given r^ and Co^ — ^01 ; required ^o, ^02 and ^oi- 

(4.) Given o^\ required To, ^02 and e^^. 

(5.) Given «5o and ^02 — ^01 ; required ro, ^02 and e^^. 

(1.) Fig. 106. If the radius r^ and the angle of advance b^ of 
the expansion eccentric are both given, the problem is readily 
solved, for then a parallelogram constructed on the main and 
expansion eccentrics (Fig. 92) will determine the virtual eccentric 



MEYER VALVE GEAR. 201 

and this (Fig. 60) the virtual valve-circle. The intersections of 
this circle with the crank positions corresponding to the limits of 
cut-off will determine the laps desired, provided the expansion 
valve does not open prematurely. In this gear, it is only at maxi- 
mum cut-off that there is any danger of an opening of this valve 
before the main valve has closed the port at the cylinder end in 
question. 

(2.) Fig. 106. Given r^ and e^^- With 0' as a center and r^ 
as a radius, describe a circumference ; the vertex D^ will He some- 
where on it. At the cut-off point i' erect a perpendicular I'Dj, 
to Oi', where it cuts the circular locus will be point D^ desired, 
provided the second intersection 2', of valve-circle on OD^ with lap 
circle e^^, is such that the expansion valve's opening does not 
occur till after the closure of the port by the main valve, other- 
wise the data must be regarded as incompatible with good run- 
ning. The lap e^^ can be found from the intersection of this 
virtual valve-circle with the crank position at minimum cut-off. 

By means of the parallelogram, the position of ODe may 
be found from OD^n and OD^ and then by method of Fig. 60 the 
position of OEg. This determines at once the setting of the 
eccentric and by applying the rule given on p. 149 the angle of 
advance do can be found if desired. 

(3) Fig. 106. Given r^ and the difference ^02 — ^01 = E. We 

know from the remarks on p. 200 that D^ must lie somewhere on 

the circumference described with 0' as a center and ro as a radius, 

and from the proposition given with Fig. 102 we know that D^ 

must also lie on a line SV, parallel to bisector OQ (Fig. 102) of 

angle (p = angle 303' = angle loi' and at a distance from it 

equal to 

_ E 

^R = ^- 

2 sin - 

2 

In general there will be two intersections of this circumference 
and line SV, which will give two solutions of the problem. That 
one should be chosen which prevents a premature opening of the 



202 MEYER VALVE GEAR. 

expansion valve. In the present case E = i ^", Vo = i '^/i" and 
(p = 54°, then SR = 1.5 14''. This gives two values of OD^, 
1.82'' and 2.20I' \ of these the former guards thoroughly against 
a premature opening for the whole range of expansion ; the 
second value of OD^ permits premature opening for nearly the 
whole range of expansion. Of these two valves, OD^^ 1.82'' 
is decidedly the preferable one and is the one chosen for illus- 
tration in Fig. 106. By the aid of the parallelogram and the 
method of Fig. 60 we can now get the position of the expansion 
eccentric OEe. 

(4.) Given ^o- This determines the location of the radial line 
OEe of the expansion eccentric relatively to the crank OC, Fig. 
106. Consequently (according to Fig. 95) the vertex D^ of the 
virtual eccentric must lie on the line 0' F^ the position of this 
line relatively to a vertical and horizontal axis through 0' being 
the same as the position of OEe relatively to a vertical and hori- 
zontal axis through 0. 

The bisector of the angle \'q2' , between crank position when 
main valve cuts off and the crank when expansion valve effects 
its maximum cut-off, will intersect line OF' in a point which is a 
limit of the possible vertices D^ on O'F. The intersections, of 
the valve-circle described on OD^^s a diameter with the cut-off 
positions, will give the maximum and minimum laps desired. 
The distance 0' D^ will be equal to the eccentricity r^ of the 
expansion eccentric. 

(5.) Given ^o and the difference ^01 — ^o., = E. As in case (4), 
we find O'F to be one locus oi D^- The construction of Fig. 102 
gives a second locus 5F of D^. The intersection of these two 
loci will give O'D^ = r^ provided D^ is located on, or to the 
left of, the bisector of the angle i'02'. Perpendiculars dropped 
from Dy on to the crank at the extremes of cut-off will then give 
^01 and 602- 

Fig. 107. The Rider Valve gear is only a particular case ot 
the Meyer. In the Rider the expansion valve plate instead 01 



RIGG ENGINE. 203 

being flat, as in the Meyer, is bent into a cylindrical shape. This 
cylinder is rotated by a governor when the grade of expansion 
is to be changed, the rotation effecting the necessary change of 
lap. The principle of this construction is shown in Fig. 107, 
where the cylinder's surface and its seat are both developed into 
a plane. Here the corresponding edges of the two ports are in- 
clined to each other but are parallel to the edges of the valve. 
When the latter is in the position shown by the full Hnes the 
cut-off = 0.5, the lap is negative and equal to 1.67''; when the 
valve has been shifted through fe into the position indicated by 
the dotted lines the cut-off is o.i and the negative lap = 0.30, 
the difference of the two laps being^^ = i yi" , The amount of 
shifting ef in the flat plate is equal in the cylindrical valve (with 
helical edges) to the circular arc through which any point on its 
surface is moved. In order that the valve may take up some of 
the wear between its face and seat, this circular seat should be 
considerably less than 180°; a value of about 100° will be found 
to be suitable. The proportions and distribution given in Figs. 
105-107 are suitable for a Rider valve gear. A Meyer gear 
can easily be arranged so as to cut off from o to ^. 



RiGG Engine, Figs. 108, 109, no. 
(See Rigg's Treatise on the Steam Engine, Pis. 17, 18, pp. 95-97.) 

In this engine the expansion is varied by a changeable eccen- 
tricity, the setting of the eccentric remaining the same at all 
grades of expansion. The centers Eg of the expansion eccentric 
must therefore lie on the same radial line OEgjEg^, Fig. 109, and 
the the center D^ of the virtual eccentric must likewise lie on one 
straight line 0' D^JD^^ whose location with respect to the axes 
X'(7F is like the location of OE^fie^ox ODefie^ to the axes 
X(9F (see also Fig. 95). 

We again assume the extremes of cut-off as given and the 
Hmits within which the expansion valve may open without per- 



204 RIGG ENGINE. 

mitting a second admission during the same stroke. The laps, 
setting and throw of the main valve are also known. As these 
data are like those for the Cummer gear the remarks on p. 194 con- 
cerning the limitations to which the locatien OD^ is subjected 
apply here also. 

We shall discuss the problems connected with this gear as 
follows : 

(i.) Given ^o, required (^o, Vo^ and To^. 

(2.) Given ^o, required ^o, r^^ and r^^. 

(3.) Given Vo^ — Vo^, required (^o, ^o, ^o, and Vo^. 

(i.) Given eo. This case is solved in Fig. 109. JO and KO are 
the crank positions corresponding to minimum and maximum 
cut-off, respectively. OH = 01=^ eo is the given, negative, lap of 
the expansion valve. Here the diameter OD^^ of the virtual 
valve-circle for minimum cut-off is taken as located at one of its 
limits, namely that one which corresponds to a reopening of the 
expansion valve at the very instant that the main valve closes. 
It would have been better to have assumed some margin. Posi- 
tion OD^^ is therefore a bisector of the angle included between 
the prolongation of crank position JO and the prolongation of 
crank position corresponding to the closing of the cylinder port 
by main valve. A second locus of D^,^ is the perpendicular at 
H to JOH] the vertex D^^ lies at the intersection of this bisector 
and this perpendicular. For this kind of gear all the points D^ 
lie on a straight line through 0\ therefore by joining D^^ and 0' 
we get a locus of the unknown vertex D^^. A second locus of 
D^^ is obtained in the perpendicular erected at / to the crank 
position KOI for maximum cut-off We have O'D^^ = To^ and 
0' D^^ = Vo^ ; the remaining unknown element b^ can be found 
by means of the parallelogram and Fig. 60, as in the preceding 
valve gears, or more directly as indicated in the figure. The 
crank is supposed to be at OX when the expansion eccentric is 
on radius OE^JE^^. 

(2.) Given Oq. This is the preceding case worked backward. 



RIGG ENGINE. 205 

We therefore use the same figure, 109. As the locus OEgfit^ of 
the expansion eccentric is given we find at once ODe^Dg^ and its 
parallel 0' D^fi^^. The latter is the locus we need and its inter- 
section D^^ with the limiting position of diameter OD^^ must be 
projected on crank position y(9// to get the lap OH = e^. With 
this lap we can now easily find ODv^. Then will 0' D^^ = r^^ 
and O'D,^ = r,,. 

(3.) Fig. 1 10. Given r^^ — ^oi- Ii^ this case we do not know 
either of the eccentricities of the expansion eccentric at the cut- 
off limits, we simply know the change of eccentricity which our 
mechanism will permit. We again assume, for the location of 
the diameter of virtual valve-circle corresponding to minimum 
cut-off, the limiting line 01, Fig. no. 

We can easily find a locus for the vertex D^^ belonging 
to maximum cut-off. For this purpose draw through 0' a 

series of lines 0' L, O'M, 0' N , which cut line 01 in 

points /, m, n, ; then lay off distances /Z, mM, nN, 

each equal to the given variation of eccentricity r^^ — ^oi. 

The curve L,M,N,D^^ thus obtained is part of a conchoid and is 
the locus mentioned. Each point on this curve represents a dif- 
ferent maximum cut-off, for the constant difference r^^ — Toi, and 
our task now is to find the point corresponding to the given maxi- 
mum cut-off. This is most easily accomplished by a trial process, 
which we will now give. Inspection of Fig. 109 for the preced- 
ing cases, shows that the intersection C of the two perpendicular 
677 and 67 dropped from the vertices D^^ and D^^, falls on the 
line 6^6^ bisecting the angle /6^ir included by the crank positions 
at the cut-off limits. This suggests the following tentative 
method : Reproduce, on a separate piece of tracing cloth, (see 
right half of Fig. no), the bisector G^ C and the perpendiculars 
CI'D" and H'Ciy so that angles G^CD'^ = GCD,^ and 
G'C'D' = GCDy^, Then place this tracing on the drawing so 
that G' C' on tracing will coincide with GC on drawing and slide 
bisector G' C on bisector GC till the intersection D^^ of perpen- 
14 



206 STURTEVANT ENGINE. 

dicular CD' with line OD^^ and the intersection D^^ of perpen- 
dicular CD" with conchoid LMN both fall on the same straight 
line 0' D^fiv^ drawn through 0' . When this occurs ail the 
conditions of this particular case are fulfilled and the size and 
setting of expansion eccentrics can be found in the now well- 
known manner. 

Before leaving this problem it should be noticed that the locus 
of the vertex D^^ (of the diameter OD^^ of the valve-circle cor- 
responding to a given maximum cut-off), is an hyperbola 
RODy^Q, whose asymptotes PL and PT d,vQ parallel respectively 
to the perpendicular CID^^ and the diameter OD^^. The origin 
O is on the one branch of this curve and 0' on the other branch. 
Points on the hyperbola are easily obtained as follows : Take 
any point Con the bisector 6^6" and drop the perpendiculars CH 
and CI on JH and KL Then prolong HC till it cuts given line 01 
in some point, say D" ^. Join this point with 0' and prolong the 
connecting line till it cuts the second perpendicular CI in D^^. 
This will be a point on the hyperbola. It is evident that the 
intersection of conchoid and hyperbola give at once exactly the 
desired point D^^. But the construction of both these curves, or 
even partial arcs of both of them, will seldom be advisable, as the 
problem can be much more easily solved by the trial method 
detailed above. 

In the preceding double-valve gears only one of the elements, 
(lap, setting and throw) has been varied at a time. In the remain- 
ing gears, two elements, or their equivalents, are varied simul- 
taneously. 



Sturtevant Engine, 12x24, Figs, hi, 112, 113. 

In this engine the expansion is varied by a simultaneous varia- 
tion of the throw and angle of advance. This variation is effected 
by a suitable mechanism, of the swinging eccentric type, the 
center of the expansion eccentric being compelled to travel on 
the circular locus Ee (Fig. 112) when the expansion is to be 



DOUBLE-VALVE GEAR, INVARIABLE DRIVING ECCENTRICS. 207 

varied. This locus passes <9 at a distance of ^''. Its center 
and radius are inscribed on the figure. According to Figure 95 
the locus of D^ has, relatively to O' y the same position as locus 
Ee to O. 

Here also we assume laps, setting, throw and distribution of 
main valve as completely determined. The range of cut-off for 
expansion valve, and the limits of the opening of this valve are 
likewise supposed to be known. On the diagram of Fig. 112 
the minimum cut-off is stated to be o. This is approximately 
true but not exactly. A tangent touching both lap circles AO 
and locus of D^ will determine, at its point of contact A, the 
crank position AOC (or least cut-off. 

To find the position of expansion eccentric at o cut-off, we 
erect at ^ (Fig. 112) the perpendicular ^Z^^,^ ; it will touch the 
locus of B^ in B^,^ and OB^^ will be the diameter of the virtual 
valve-circle. The dotted parallelogram OD^fie^D^t gives the 
relative position and magnitude of the diameters of main, expan- 
sion and virtual valve-circle ; the equal parallelogram OE Eefim 
gives the relative position and magnitude of the corresponding 
eccentrics. OEg^ = Tq, is the position of the expansion eccentric 
relatively to crank OC when cut-off = O. 

Fig. 113. In like manner we find that 0E>^^ is the diameter 
of the virtual valve-circle for the maximum cut-off = 0.7. The 
reopening of the port by expansion valve occurs just after main 
valve closes. For this cut-off the position and size of expansion 
eccentric is given by OEe^ when crank is at dead point OC. The 
expansion eccentric therefore travels on locus from OEe^ to OEe^ 
while varying the expansion. 



Double- Valve Gears with Invariable Driving Eccentrics. 

There are two kinds of double-valve gears in which expansion 
valves are driven by non-adjustable or invariable eccentrics. In 
the first kind the expansion valve is driven by one or more in- 
variable eccentrics, and its absolute motion is entirely indepen- 



208 GUINOTTE VALVE GEAR. 

dent of the main eccentric. In the second kind the absolute 
motion of the expansion valve does depend, more or less, upon 
the main valve's eccentric. In both kinds there is between the 
expansion valve and the invariable eccentric a mechanism whose 
action is equivalent to driving this value by a variable eccentric. 

A valve gear by G. Schuhmann, of Reading, Pa., belongs to 
the first kind. Between the expansion eccentric and its valve 
there is a rocker with two arms, a variable and invariable one, 
the invariable one driving the expansion valve and the variable 
one receiving its motion from the expansion eccentric through a 
block whose position on the arm is regulated by the governor. 
The locus of the center Eg of the expansion eccentric is a straight 
line which, when crank is on back center, passes below and to 
the right of the shaft center making an angle of 30° with the 
line of dead centers. The main valve in this engine belongs to 
the positive indirect, and the expansion valve to the negative 
indirect, type. 

The valve gears of Polonceau (or Borsig), of Guinotte and of 
Bileram are examples of the second kind. Polonceau's is de- 
scribed and discussed on p. 203, second English edition, of 
Zeuner's valve gears. Here both main and expansion valves 
are driven by Gooch's link motion and the locus of the center 
E^ of the virtual eccentric is a vertical line through center of 
shaft. Of these gears we will discuss only those of Guinotte 
and Bilgram. 



Guinotte's Valve Gear, Figs. 114, 115, 116. 

The complex mechanism and theory of this gear are both 
given by Dr. Zeuner (see p. 214, second English edition of Valve 
Gears). It permits the reversal of engine without touching the 
expansion gear. The character of the locus of the vertex of the 
diameter of the virtual eccentric is such as seems to deserve 
special treatment here. We therefore reproduce it although it is 



GUINOTTE VALVE GEAR. 2O9 

fully given by Dr. Zeuner. In this gear the vertex D^ of the 
diameter of the virtual valve-circle lies in a rectilinear and in- 
clined locus iWV, Fig. 115. This is similar to the locus MM o^ 
the corresponding vertex De of the expansion valve-circle and 
this in turn corresponds to locus LL of the center of the equiva- 
lent expansion eccentric. We say equivalent because the actual 
expansion eccentric has a constant eccentricity OD^-=^ro and a 
constant angle of advance, ^0=90°. But the mechanism be- 
tween it and the expansion valve has the effect of moving the 
latter as if it were driven by an expansion eccentric whose center 
had a locus LL for the different grades of expansion. Accord- 
ing to the principle underlying Fig. 95, the position of NN rela- 
tively to O as an origin is the same as that of iM/ relatively to 
O as an origin and the same as LL relatively to 0. 

At position 3 (Fig. 115), the main valve cuts-off at the left 
cylinder port and in order that the expansion valve shall shut its 
left (main valve) port at the same time we must give the diam- 
eter of the virtual valve-circle a position OD^^, which is found by 
erecting, at the lap-circle /, the perpendicular LD^j^ to the cut-off 
position lO\. The distance OD^ will be the corresponding 
diameter of expansion circle OEe^ the corresponding position 
of the equivalent expansion eccentric. In the mechanism itself 
this corresponds to the position of the block at the upper end of 
the link. It is evident that with negative lap the virtual valve- 
circle OIDy^ will cut-off at the same instant that the main valve 
closes. 



Fig. 1 16. For a minimum cut-off we will choose o or cut-off at 
dead point. To find the corresponding virtual valve-circle we erect 
to the dead center line, at the lap circle H, the perpendicular HD^^ 
and prolong it to the locus NN of the vertex D^^. The dis- 
tance ODy^ will be the diameter of the virtual valve-circle 
OHDy^. It is evident that this gives the desired cut-off and that 
the reopening of the left port in the main valve will not take 
place till crank reaches position 6^4, that is after the main valve 



210 BILGRAM VALVE GEAR. 

has closed the left cylinder port. The position of the equiv- 
alent expansion eccentric will be near ODg^ in this case and in 
the mechanism it will be found that this requires that the block 
shall be at the bottom and not at the central portion of the 
curved slot in the link. 



BiLGRAM Valve Gear, Figs. 117, 118, 119. 

The mechanism of this gear has been given on p. 119 of Bil- 
gram's work on Slide Valve Gears, also in Halsey's Valve Gears, 
p. 123. There is only one eccentric which drives the main 
valve directly and the expansion value indirectly. 

The inventor gives the favorable porportions, for most of which 
the reader must look to the references just given. The mechan- 
ism between the main eccentric and the expansion valve, in its 
action on the latter, is nearly equivalent to the action of an 
expansion eccentric whose center travels on the circular 
locus OLEe, Fig. 118, while passing from one extreme of expan- 
sion to the other. A line OE^ = 2 X OE,n = 2r placed 
50° in advance of the main eccentric radius OE^ is the diameter 
of this locus. The corresponding circular locus OMD^ of the 
vertex D^ of the diameter of the virtual valve-circle is found in 
accordance with the principle illustrated in Fig. 95. 

Fig. 118. The inventor recommends a negative lap for expan- 
sion valve that is *\ than the outside lap of the main valve. We 
have here taken y^" for this negative lap because this is favorable 
to quick cut-off. To get a cut-off at the dead point erect to 
OH the perpendicular HD-^^ and prolong it till it cuts the locus 
OMD^fi^^ in D^^. Then will OHD^^ be the required virtual 
valve-circle for minimum cut-off It is evident that the reopen- 
ing of the left, main valve, port, occurs at position 4(9^ after the 
main valve has closed the corresponding cylinder port. 

Figs. 118, 119. The cut-offs at one-quarter and one-half 
stroke are treated in the same way; the corresponding virtual 



PRACTICAL DATA AND DIRECTIONS. 211 

valve-circles are OID^,,^ and OJD^^ (Fig. 119)- Inspection of 
these circles at the points / and J shows that the cut-off is very 
quick, as is claimed. 

The range of cut-off is great, extending from nothing up to 
that effected by the main valve. The virtual valve-circle corre- 
sponding to this maximum cut-off is OKD^, Fig. 119. 



Practical Data and Directions. 

The rapidity of cut-off for any gear depends upon the speed 
of the valve at the instant and the number of ports through 
which the steam is admitted to one end of cylinder. It can 
easily be estimated by finding the position of the virtual eccen- 
tric at this instant and dropping a perpendicular from its center 
upon the " dead point " line of the eccentric. The product of 
the number of ports and the ratio of this perpendicular to the 
perpendicular dropped from crank-pin on piston stroke is, other 
things being equal, perhaps as fair and convenient a measure as 
any for comparing the rapidity of cut-off of the different gears.* 

The distance from the port of a valve edge not engaged in 
steam distribution should be more than sufficient to prevent its 
opening the port at maximum travel. 

To avoid two admissions of live steam during the same stroke 
the expansion should not reopen its port in the main valve till 
after the main valve has closed the corresponding cylinder port. 

Arrange the valve gear so that it will realize the steam distri- 
bution required by the desired indicator diagram. The release 

* A more exact measure would be the product of the number of ports by 
the ratio of valve speed to piston speed at instant of cut-off. The valve 
speed is measured by the intercept included between the shaft's center and 
the eccentric-rod, the intercept being taken on the perpendicular through 
this center to the stroke of the valve end of the rod. The piston speed, to 
the same scale, is measured by the intercept on the perpendicular to its stroke 
included between shaft center and connecting-rod. The ratio of valve speed 
to piston speed is the ratio of these two intercepts. 



212 PRACTICAL DATA AND DIRECTIONS. 

may begin at if of the stroke, unless it has already been fixed 
by the compression assumed in the earher part of the work. As 
American practice now is to use balanced valves, large throws of 
eccentrics are permissible. The cut-off should be made as rapid 
as possible. 

The width of bridge (metal between steam and exhaust ports) 
should be greater than r — (^ + ^) to avoid fresh steam passing 
directly into exhaust. It is often made equal to the thickness 
of the cylinder wall. An empirical formula given by Zeuner is 
0.4 + 0.5^. 

To prevent steam leaking between valve face and seat they 
should have a contact-width of at least 0.4 of an inch. 

The exhaust port should have such a width bo, that when 
valve is at end of travel, it will still have the width b, of the 
steam port, that is, ^o = ^ + ^ -f 2" — (0.4 -|- 0.5^^) 
= r + o.^b + 2 — 0.4. 

The valve seat should have such a length that the valve will 
pass beyond the outer edges of the seat, thus avoiding the for- 
mation of a shoulder in the seat. These should always remain, 
however, a 0.4" contact to prevent leakage. 

When cut-off £, lead v and maximum port-opening b ± k 
(-j- /^ = over-travel, i.e., beyond edge of port) are given, the 
half-throw r can be found by trial, or at once and exactly, by 
Zeuner's formula : 



_2 {b ±. k) — V ^ 2 V{b d= k){b ± k — v){i — s) / \ 

2£ 

The following tables will be useful in the solution of problems 
relating- to the main valve. When the outside lead is known as 
a fraction of the eccentricity r, for example v = mr, we may 
transform the formula into 



b -\- k ^ m \ in^ m^ 



2 \ 4 4£ 



PRACTICAL DATA AND DIRECTIONS. 



213 



Values of 



b ^k' 



Values 
of 

V 

— = m 
r 

0.05 
0.075 
0.10 
0.125 


Values of t = apparent cut-off. 


.50 


•55 


.60 


.65 


.70 


.75 


.80 


.85 


.90 


3.140 
3.018 
2.906 
2.792 


2.821 

2.721 
2.626 
2.536 


2.545 
2.464 
2.387 
2.314 


2.307 
2.240 
2.176 
2.116 


2.094 
2.039 
1.987 
1.937 


1.905 
1.859 
1.815 
1-774 


1.730 
1.693 
1.657 
1.622 


1.567 
1.537 
1.508 

1.479 


1.410 
1.386 
1.362 
1.339 



Values of - 
r 



Values 
of 

V 

m = — 
r 

0.05 
0.075 
0.10 
0.125 






Value of e = 


= apparent cut-off. 






0.50 


0.55 


0.60 


0.65 


0.70 


0.75 


0.80 


0.85 


0.90 


.681 
.669 
.656 

.642 


.645 
.632 

.619 
.606 


.607 

•594 
.581 
.568 


.566 

.554 
.540 
.527 


.522 
.510 
.497 
•484 


.475 
.462 

.449 
.436 


.422 
.409 

•383 


.362 
•349 
.337 
.324 


.291 
.278 
.266 

.253 ! 



For any assumed value of 



b-^ k 



we can immediately get the 



angular advance from 



• > e -^ V 

sm O =z \ :^ I 



-J — -I- -= I + m 
r r 



b^-k 



(94) 



k should be taken as small as is consistent with sufficient port 
opening. The shorter cut-offs given in table are more favorable 
than the longer ones in preventing the reopening of the expansion 
valve before main valve closes, on the other hand the shorter 
cut-offs involve a greater throw of main valve. 

In slow going engines the outside lead is often taken = tV r, 
while in fast running engines ys r is often taken. 



214 VALVE GEARS WHICH DO NOT POSSESS HARMONIC MOTION. 



Valve Gears Which do not Possess " Harmonic " Motion. 

On p. 1 34 we divided all valve gears into two classes : those 
in which the valves move approximately like the " slotted cross- 
head " and those which did not, that is, into valves with and: 
without " harmonic " motion. It is the first class that is most 
extensively used in practice and the one which has been dis- 
cussed in the preceding pages. Even in these there is some 
deviation fi-om " harmonic " motion which is not due to the 
angularity or length of the eccentric rod. As the valve stem is 
often at a considerable distance from the plane of the eccentric, 
sometimes at the side, and at other times on top, of the cylinder, 
rocker arms are interposed between eccentric and valve to trans- 
mit the motion. Usually they produce less deviation from "har- 
monic" motion than the angularity of the eccentric-rod. But 
they may be arranged so as to produce a rapid opening of the 
port followed by a period of comparative rest for the valve. The 
Corliss valves driven by a wrist-plate and the Porter- Allen valves 
driven by the bent lever O'R, PQ in Figs. 80-82, are examples. 
Mr. F. A. Halsey has shown, in his Slide Valve Gears, pp. 54-63: 
and 89-98, that the rocker arms may be arranged so that both 
cut-off and lead are equalized. In the Ide engine they are used 
to keep the lead constant at each end but greatest at the cylinder 
end farthest from shaft. All such devices produce some devia- 
tion from "harmonic" motion. 

The most characteristic members of this group are the so- 
called "Radial" valve gears, of which the Joy is an excellent 
representative and perhaps the one which is best known in this 
country. The link driving the valve rod in these gears has two 
of its points guided in curves, one of which is closed and the 
other open or closed, while a third point drives the valve rod 
or stem. The motion is usually of a complex character and the 
valve travel is got at by plotting the paths of the three points of 
the driving hnk, in much the same fashion as the travel of the 
Porter-Allen valves was obtained. The relation between valve 



VALVE GEARS WHICH DO NOT POSSESS HARMONIC MOTION. 21 5 

and piston travel may also be shown by polar or oval diagrams 
like those given Figs, 83-91 for the Porter- Allen valve gear. 

This group of non-"harmonic" valve motions (see p. 134), has 
these two following sub-divisions : 

(a.) Gears in which the valve slides on a stationary seat and 
only its absolute motion is considered. 

(d.) Gears with two valves, of which one slides on the other, 
and in which the relative as well as the absolute motions of the 
valves must be considered. 

The Porter-Allen motion, already discussed, belongs to the 
first sub-division and may be regarded as composed of three 
"single-valve" gears, each valve sliding on a fixed seat. 

The Payne-Corliss engine belongs to the second sub-division. 
It has two sets of double-valves, one set for each end of the cyl- 
inder. Between the eccentric and the valves there are Corliss 
wrist-plates, but not the Corliss releasing-gear. 



2l6 CRANK ANGLE AND MAXIMUM VELOCITY OF SLIDE. 



APPENDIX A. 

Crank Angle Corresponding to Maximum Velocity 
OF Slide. 

Dr. C. L. Schadwill in a thesis presented in 1876, and entitled 
"Das Gliedervierseit," proved that the configuration of the sHder- 
crank giving zero acceleration (or maximum velocity) of the slide, 
see Fig. 1 20, existed when line of instantaneous centers 1 3-24 



F,g. 120 




CRANK ANGLE AND MAXIMUM VELOCIXY OF SLIDE. 21/ 

is perpendicular to rod's direction. (It has occasionally been 
erroneously assumed that this configuration existed when line 
13-23 is perpendicular to rod.) 

Starting with this proposition we will now find a cubic equation 
for the relation between this particular crank-angle and the 

ratio — of crank to rod. 

Figure 120 readily furnishes the relation. 

OW 



OWtan = OW {t3in /9 -|- cot /9) 



sin /9 cos ^ 
or cot (^ =: sin /5 cos j3 

and since sin ^ : sm =^ r \ I 

we get, after reduction, 

sin^ Q ^ — sin4 Q sin^ Q = i. 

In terms of the rod angle /9 the cubic equation takes an even 
simpler form, 

sin^ /5 H- sin^ /9 — sin^ /3 = — . 

In either of these equations we have an exact solution. 
Grashof in his Machienenlehre, Bd. II, p. 133, found approxi- 
mately, 

r fr\^ 

cos0=--ty , 

which is correct to within ( - ) * 

This result is of course sufficiently exact for engineering 
purposes. 

V 

For - =r= ?r. cos Q = 0.1597 and = 80° 49'. 



2l8 



CRANK ANGLE AND MAXIMUM VELOCITY OF SPEED. 



ts 


::::. 


*«s^ 






















^ 




tk 












bL«- 




7, 


"^ 


^^^^^^>^n^ 




154 -4a 
■n 






-4 


o~ 


15 


30 45 

Ab'-lZ 


60 


7s c->^y-v^.os 


120 


i;v5 


150 


(65 


^ 






Fig 121 






^^*S! 


d 


■^ 


''^ 







On pages 36 and 37 it is shown that the sHde acceleration is 
approximately, 6 = - \ cos ± - cos 2 \- This means 
that ordinary finite rods have same velocity as an infinite rod 



when crank angle is about 45°. Thus, for — = i, this angle is 

^.= 45° 12' or 134° 48'. 

This gives another convenient method of constructing approx- 
imately the curve of slide accelerations abcde, Fig. 121. For we 
have for the infinite rod the straight line fbmdg. For the finite 
rod we have exactly af =^\hf and eg =^\ ek, and approximately, 
there is equality in the acceleration of the two rods at bl and nd. 
Point c we can find closely from Grashof s formula given above. 
This enables us to get easily five points abcde on the curve of 
sHde accelerations and leads to a ready and sufficiently accurate 
construction of this important curve in Steam Engine Dynamics. 



INERTIA- RESISTANCE OF THE ROD. 2ig 



APPENDIX B. 

Diagram for Determining Inertia-Resistance of the Rod, 

THE Horizontal and Vertical Components of this 

Resistance and the total Force 

Exerted by Rod. 

The method of finding the resultant of the weight of the rod 
and of its resistance to change of velocity has been given on 
page 103, Figure 47, and the method of finding the pin pressures 
is illustrated on page 108, Figure 49. In both the figures, the 
main train of engine (crank, rod and slide) is supposed to be 
drawn to some small scale to bring it within the limits of the 
paper. Even then, many of the construction lines are long, and 
separate, enlarged, diagrams of the polygon of forces WVbMs, etc., 
are found necessary. These disadvantages can in large measure 
be removed by omitting the connecting-rod from the drawing, then 
using in its place the image Civ (see figures 43 to 46) of the rod, 
and making the constructions employed with the image, similar 
to those used with the rod. The loUowing figure, 122, illustrates 
how this may be done. 

Cgwz is the image of the portion C<& WZ of the rod. Triangle 
jgfz is the image of triangle J^ZF and is the triangle /'^y*'^^; 
these three triangles are all similar to each other by construction. 

The pointy through which passes the resultant of rod's inertia 
resistance is obtained by finding that point Z of rod which has 
acceleration along the axis of rod and using it as one point of 
mass concentration, the other point J of mass concentration 
being taken so as to satisfy Eqs. 68 to 70 on p. 95, that is, so that 

This will necessitate the rod's inertia-resistance passing through 
point /. 



220 



INERTIA-RESISTANCE OF THE ROD. 




INERTIA-RESISTANCE OF THE ROD. 221 

Point Z is horizontally projected on rod from point z of image 
which point in turn was found by drawing Oz parallel to the 
center Hne CW of the rod. When rod is omitted from the figure, 
point z is projected to z" on crank OC and the right-angle 
triangle z/y is used to determine/' (the horizontal projection of 
pointy.) Here the perpendicular 

The point/' can be laid off on a circle of its own, its zero line for 
angles being along line O/o. All points /' are easily found from 
the y and/' points of the figure. We will assume, without again 
giving the construction, that the/' points of Fig. 122 have been 
correctly found and then determine the ^ points of division on 
Og: Then will g^- and o-Q respectively represent the components 
at wrist-pin and crank-pin that make up, and are parallel to, the 
whole force of acceleration of rod. 

To find the resultant of this force and the weight we may pro- 
ceed as in Fig. 47, page 103, but this involves drawing the 
crank train. Or on (9^ and OL (here the weight OL is, for conve- 
nience, taken very large relatively to the inertia O^) we may con- 
struct the parallelogram OgML and OM'is the resultant desired. 
By laying off the weight OL in an opposite direction, 00' ^ and 
completing parallelogram OO'gMv^o. can get the desired resultant 
O'g as the upper side of the parallelogram. It is the last con- 
struction which is most convenient and is the one used on 
plate. 

We next resolve this force 0'g-=^ OM into its actual, com- 
ponent, pressure at wrist-pin and crank-pin. This is done as in 
Figs. 48 and 49, the distance O'g (or OM) Fig. 120, taking the 
place of WZ, Fig. 48 or of WM, Fig. 49 ; and the point 0' 
(or 0) then taking the place of point W in Figs. 48 and 49. 
The first step is to find the point 5 on resultant which will divide 
it into the two parallel components SO and SM {SW and SZ, 
Fig. 48) acting at wrist-pin and crank-pin respectively. It is 
evident from equation 72, page 107, that the point S must lie on 
15 



222 INERTIA-RESISTANCE OF THE ROD. 

a line QS {SVt Fig. 48 or SZ Fig. 49) which is parallel to 
the line of centers WC and passes through the intersection Q of 
any pair of components of OM, for instance OQ and QM. It is 
evident that we shall know Q if either component OQ or QM is 
completely known. In the figure we have resolved the weight 
OL into components (9Tand TL parallel to itself and acting at 
wrist- and crank-pin respectively, the components satisfying the 
condition OT\TL^= C^:%W. In like manner the resistance 
to inertia is resolved into the two components Eg and EO, satis- 
fying the proportion Eg : EO = CJ '.JW. 

Transferring ^^ to (97V and combining latter with OT we get 
OQ, one component of OM. We now draw through Q the Hne 
QS parallel to line of centers CW of rod, and then through 5 
the line Sb (see also Fig. 49) parallel to line of internal stress, 
i. e., parallel to the proper tangent to the friction circles. In 
the figure extravagantly large friction circles have been placed at 
(7 and ^in order to show the construction more clearly. 

In the figure on plate the forces are transferred from center of 
shaft to point 0' above, the latter at distance 00' -— OL =: 
weight per □'' of piston. Then (Figs. 122 and 1 24) O'g = 0M=^ 
resultant of weight and inertia-resistance of rod. Furthermore 
it is evident that this resultant gO' can be divided at S' into the 
same segments as OM at S, by laying off NQ' = TL and draw- 
ling Q' ^' parallel to center line CW of rod. 

Here it will be well to tabulate the value of component 0' S' in 
some table of forces ; it can then be conveniently used in the 
diagram for finding crank- and wrist-pin pressures. In the 
present example, the values of 0' S' are : 



20° 


40° 


60° 


80° 


100° 


120° 


140° 


160° 


180° 


1-59 


1.68 


1. 10 


0.60 


0.39 


0.00 


-0.90 


-4.41 


5.21 


200° 


220° 


240° 


260° 


280° 


300° 


320° 


340° 


360^ 


2.21 


1.54 


1.39 


1.28 


1.40 


1.91 


2.61 


3-59 


6.00 



The minus sign indicates that point S' has fallen outside of 
space O'g. 



CONSTRUCTION OF DIAGRAM OF INERTIA-FORCES OF ROD. 223 

In order to construct curve aaaa in the diagram of shaking 
forces, Fig. 50, we must know the horizontal and vertical com- 
ponents of the inertia-resistance of the rod. By projecting the 
pointy (Figs. 122 and 124) to g' , on the vertical through 0' we 
get in gg' the horizontal, and in Og' the vertical, component of 
the inertia-resistance Og of rod. The values of these components 
should be tabulated in the table of forces shown with Fig. 50. 
They can then be easily transferred to the diagram for finding 
wrist- and crank-pin pressures ; this is particularly desirable if 
scale of the pin-pressure diagrams differs from that of the rod's 
inertia-diagram. 

Construction of Diagram of Inertia-Forces of Rod. 

For ease of construction and accuracy it is well to make the 
crank of diagram about 12 inches long. The exact length of 
crank is determined by scale of forces, which should be a con- 
venient one. In this case i pound per \^" of piston area to the 
linear inch will be a convenient scale, making the radius 

for in this 14x14 engine represented in the diagram, A = 153.94, 
W=z 136 pounds and revolutions per minute 273, while 
weight of reciprocating parts is 321 pounds. Take the crank 
positions 20° apart and then find the foot w of the image by 
computing the distance Ow by means of table on page 39. 
We have 

Ow = 13.02 X tabular quantity c. 

The rod is supposed to be six cranks long. 

When the segments O'S^ and S'g of the total force O^g of the 
rod are to be found for each one of a series of crank positions, the 
work of locating on the image the point g corresponding to the 
center of gravity of the rod can be greatly simplified by dividing 
crank-radius OC of diagram. Fig. 124, at pointy' into the seg- 

ment Og' so that-^^ equal ratio -jz^'m rod, (see Fig. 122), and 



224 CONSTRUCTION OF DIAGRAM OF INERTIA- FORCES OF ROD. 

then describing about a circle with the radius Og' . In Fig. 124 
this radius is 0.573 of the diagram crank. Where this circle cuts 
the crank positions draw horizontal Hnes to point g of images 
corresponding to these positions. The lines Og represent the 
total inertia-resistances offered by rod at these crank angles. 

When the crank is upon its dead points the method followed 
above fails because then the image coincides in direction with the 
radius. In this case the point g on image may be found by 
making, for the crank-angle 0°, the distance 

R\ ^ , C(B> 
L 



Og =1 R f I ± ip --\ when (p = — — , (Fig. 122); 



here C(3 is the distance of the center of gravity from crank-pin 
center. 

The reason for this construction is that wrist-pin W's accelera- 
tion ats 00 [is, in our diagram, measured by i?( I ± -—jrespec- 

tively. Here — - represents the length of the image to our scale. 

As the g point always divides the image into segments having 
the same ratio as the segments on each side of center of gravity 
of rod, that is, in the ratio (p, we get, Og = radius db ^ X image 
as above. 

To obtain inclination a of rod readily we can compute it from 

sin a =-— sin co, or, we draw in lower half of diagram a semi- 

circle having a radius i that of diagram crank and horizontally 
project this circle's intersections with the crank on to the crank- 
pin circle. The radii joining these projections with the center 
will evidently have the inclination of the rod corresponding to 
the crank positions. 

We have alread}^ described, with the help of Fig. 122, the steps 
that must be taken to divide the total force O'g of rod into two 
parallel components S'O' and S'g acting at wrist- and crank-pin 
respectively. It is therefore unnecessary to repeat it with present 
figure as these two figures are lettered alike. In Fig. 124 the 



STEPS, STATED IN THEIR ORDER. 



22: 



wrist-pin pressure S'O' is not drawn, for the sake 
of clearness; its value can however be taken directly 
from diagram by measuring from fixed point 0' to 
end S' of the segment gS' that is drawn. At the 
dead points (0° and 180°) the point E on line Og 
representing inertia-resistance becomes indeter- 
minate, but the resultant O'g of inertia and weight 
of rod can nevertheless be resolved into its two 
parallel components at crank and wrist-pin. First 
it is evident that this resultant cuts the line of rod 
at the center of gravity itself and its components 
at the pins will be inversely as the distance of the 
pins from the center of gravity. The curve con- 
necting the points S' is very irregular, has a loop, 
and has four branches running out to infinity. For 
the present data its course is as in Fig. 123. 

The following enumeration may be of assistance 
in expediting the work of constructing the diagram. 



Steps, Stated in Their Order. 

1. Draw crank positions for every 20° of rotation. 
(Additional positions may subsequently be added to 
fill out parts of other diagrams of special interest.) 

2. Draw three concentric circles, for point of 
concentration, center of gravity and angle of rod. 

3. Draw images of rod, using table on page 39. 

4. Project horizontally the intersections, of crank 
with circles (mentioned above in 2), on to image 
and on to crank-pin circle. It is evident that the 
projecting lines will be the same for symmetrical 
crank positions in first and second quadrants. 

5. Draw parallels to rod, as in diagram of rod 
angles. Fig. 124. 

6. Lay off weight OG upwards from center 
and divide it into the two components that act at 
crank and wrist-pin. 



226 DIAGRAM OF PRESSURE AT CRANK- AND WRIST-PIN. 

7. Draw direction of acceleration of center of gravity by join- 
ing center of shaft with point g. 

8. Draw the line Oz parallel to direction CW of rod till it cuts 
in z the prolongation Cwz of the rod's image. Through this 
intersection draw a parallel zz' to stroke of l^till parallel cuts in 
z' the crank CO or its prolongation. 

9. On the hypothenuse Cz construct the right-angled triangle 
c'g'ff similar to Z%JF. A simple way will be to make perpen- 
dicular 

CO P 

^^' = ^' = ^-^=1^= 

this need only be done once. With 6^ as a center and Of as a 
radius describe a circle ; this will be the locus of all the/' points 
of the series of crank positions. 

10. Having found/' as directed by 9, draw the parallel j'q to 
image, cutting acceleration of center of gravity at q and dividing 
the inertia-resistance Og into two segments. 

1 1. Lay off ON, from center 0, equal to ^^the smaller of these 
segments, on this same Hne of acceleration of center of gravity ; 
then lay off upward the greater of the gravity components. 
These two components can be laid off at same time on, and from, 
the symmetrically placed Og line in the lower semi-circle of dia- 
gram, Fig. 124. 

12. Through the upper end of the greater, gravity, components 
draw lines of internal stress parallel to inclination a of connecting 
rod, (see inclination lines in lower half). 



Diagram of Pressures at Crank- and Wrist-Pin. 

Figure 125. 

The preceding diagram gives the total force O^g exerted by 
the rod, its segments S'O' and 6"'^ representing those components 
of the pin-pressures that are parallel to total force of rod. If we 
combine the internal stress (see Fig. 49 and page no, line 23) 
with each of these components we get the actual pin-pressures. 



DIAGRAINI OF PRESSURE AT CRANK- AND WRIST-PIN. 22/ 

The wrist-pin pressure thus formed must be equal to the resultant of 
the guide reaction and of the pressure acting along axis of cylinder. 
This pressure is the algebraic difference between the effective 
pressure of steam and the inertia-resistance of reciprocating parts. 

The principle underlying the construction of these pin pressures 
has already been fully explained in connection with Figs. 48 and 
49. We shall therefore confine ourselves to stating how the 
procedure may be simplified and drawing-paper economized. 

In the first place we must, in this diagram, represent our forces 
to a smaller scale than in the preceding diagram of rod forces ; 
for the pressures along axis of cylinder are much larger than the 
rod forces. Generally one-half or one-third of scale used in first 
diagram will answer. We will choose the latter reduction for 
Fig. 125, that is, we shall assume that in this figure the scale of 
forces is 3 pounds per □'' of piston area to the linear inch. To 
lay off exactly, in direction and intensity the total force WM of 
rod in the new diagram, we divide by 3, the tabulated value of its 
rectangular components, and lay them off on the proper one of the 
four pairs of axes we shall use. Each of these four pairs of axes 
corresponds to a quadrant of the crank's motion. As the cylinder 
is supposed to be to the right, the axial pressure will be to the 

< . 1 > for the ] , > two quadrants and the axis of X is 

made to extend to the< ,^r. > of origin W. 

The following tables give the values of the rectangular com- 
ponents O'g-^ and £■£■' for the example illustrated in the present 
diagrams. 



228 



DIAGRAM OF PRESSURES AT CRANK- AND WRIST-PIN. 



Vertical Components of Total Force of Rod per Q'^ 

-f AND — Respectively Represent Up and 

Down Components of Rod. 









Crank Angles- 


-Forward Stroke. 






o 

1 80 


10 
170 


20 
160 


30 
150 


40 
140 


50 
130 


60 
120 

5.62 


70 

JIG 


80 
100 


90 
90 


—0.88 


0.40 


1.67 


2.86 


3-93 


4.87 


6.18 


6.52 


6.72 








Crank Angles- 


—Return Stroke. 






180 

360 


190 
350 


200 
340 


210 
330 


220 
320 


230 
310 


240 
300 

-7.38 


250 
290 

—7.94 


260 
280 


270 
270 


--0.88 


-2.16 


—343 


—4.62 


—5.69 


—6.63 


—8.28 


—8.48 



Horizontal Components of Total Force of Rod per Q'^ 

-|- AND — Represent Force Acting to the Right 

and Left, Respectively. 



Crank Angles — First and Fourth Quadrants. 




360 


10 

350 


20 

340 


30 
330 


40 
320 


50 
310 


60 
300 

6.11 


70 
290 

3.79 


80 
280 

1.40 


90 

270 


14.02 


1377 


13.00 


11.80 


10.21 


8.25 


— 093 


Crank Angles— Second and Third Quadrants. 


100 
260 


no 

250 


120 
240 


130 
230 


140 

220 


150 
210 


160 
200 


170 
190 


180 
180 




-3.14 


-5.18 


-7.02 


-8.58 


-9.89 


-10.87 


-II .61 


-12.04 


-12.17 



Having laid off the total force of rod to the assumed scale 
we next lay off from the origin W, the sm.aller 0^5', Fig. 1 24 (or 
WS of Fig. 49), of the two segments of the total force and 
through the point 5 thus found draw the line Sd (Fig. 49 and 
Fig. 122) parallel to the line of internal stress, which is very 
nearly a tangent to the friction circles at crank and wrist-pin. 
These tangents are drawn, in the middle of the diagram, Fig. 
125, for the inclinations of connecting-rod corresponding to the 
different crank angles. The proper tangents for each position 



DIAGRAM OF PRESSURES AT CRANK- AND WRIST-PIN. 229 

are decided upon according to the method indicated in Figs. 38 
to 41, pages 89 and 90. 

Having drawn, Fig. 125, the line Sd parallel to the proper tan- 
gent, we next lay off on each axis of Xthat force WV, along axis of 
cylinder, which equals the difference of effective steam pressure 
and inertia of reciprocating parts. From its end Flay off the 
guide's reaction Vd, incHned, from the normal, against the slide's 
motion, an amount equal to the angle of friction. This reaction 
will determine point d and thus determine the actual wrist-pin 
pressure dW a.nd the crank-pin pressure dM. 

Of course the total labor of taking these various steps is con- 
siderably diminished by taking the same step for each of the 
crank positions before proceeding to take the next step in the 
series. In laying off the total force of rod it may be of some 
help to notice that the vertical components of this force are alike 
in quadrants I and II for symetrically placed crank positions and 
are also alike in III and IV for such crank positions. 

It is evident from the figures corresponding to the different 
quadrants that friction has very little influence on the intensities 
of the pin-pressures. It does influence the vertical component 
of the wrist-pin pressures to a notable degree because these are 
small any way. As it is just these components that determine 
whether slide shall press upward or downward and as a very 
slight upward component will suffice to effect this reversal of 
pressure and cause a knock, we cannot say that the friction's 
influence is everywhere practically unimportant. Moreover we 
must not conclude, because pin pressures are not appreciably 
affected by friction, that the latter's influence on engine friction 
is likewise inappreciable. The crank-pin's pressure for instance 
makes its influence felt to a notable degree by acting tangent to 
friction circle in such a way as to diminish its crank leverage by 
the radius of the friction circle. 

When it is only a question of finding when reversal of pressure 
on slides occurs, the determination of pin-pressures will not be 
necessary for every crank position of Fig. 124. Inspection of 



230 DIAGRAM OF PRESSURES AT CRANK- AND WRIST-PIN. 

Fig. 124 will show where this reversal is liable to take place and 
the pressure determinations of Fig. 125 may then be confined to 
the crank positions of that vicinity. 

In each of the upper and lower semi-circles, the images are 
exactly parallel for crank positions symmetrically situated with 
respect to the 90^-270° line. 

This parallelism can be shown as follows : Let r represent the 
angle CwW mside by image with stroke of sHde. Then will 

wH R cos — d 

cot r = -—- =z "r^ 

^^ R sin 



cot r = H- 



sm 



— sin^ G — j^ cos^ —1 



if- 

Here d has the meaning given it on page 37 and its factor — has 

R 

the value R in diagram, Fig. 124. From this it is evident that 

cot T will be the same for any two supplementary angles Q and 

angles 180° — Q. 

When cot r = o, the image becomes vertical ; this is always 

so for infinite rods and for common ratios of — when crank-angle 

R 

Q is nearly 45°. 

With the help of this expression for cot r, we can show 

01 = iL (cos Q -\- ssxi Q cot r) exactly, 

and that the distance 01 is practically a constant quantity for 
ordinary rods. This would enable us to find still another easy 
method of getting the location of the inertia-resistance of the 
rod ; but we omit it as it is not any simpler than tne method we 

now have. See note on page 100. ^ = —-- , Fig. 45. 



n^. 126 





D a'.- -" 

1 Xl 




l^. 1 




w 6' 




Fig \Z1 


'/x 














1 

\ / ' 

\ / 



Fig-]28 



^, 



y 




FORCES ACTING IN AND UPON THE CONNECTING ROD. 23 1 



APPENDIX C 

Exact Construction for the Forces 
Acting in and upon the Connecting Rod. 

by r. c. h. heck, m.e., 

Instructor in Mechanical Engineerin§^^ Lehigh University. 

1. When the forces developed in the connecting-rod, due to 
its own mass and motion, are not considered, the rod is taken as 
simply a connecting link, serving to transmit force from one pin 
to the other. The pin-pressures, equal in intensity and opposite 
in direction, act along the same straight line. This force Hne, or 
line of internal stress in the rod, passes through the pin centers, 
friction being neglected. Taking account of friction, the line is 
tangent to the friction circles. But in any actual engine, there 
are certain forces developed in the connecting-rod, due to gravity 
and inertia, generally acting transverse to the center line of the 
rod. It is therefore necessary to discuss what takes place under 
these conditions. 

2. In Fig. 126, let WC represent any rigid body, upon which 
acts the force Q along the line XZ. Impose the condition that 
the body shall be held in equilibrium by two forces passing 
through the points W and C respectively. These two must 
intersect in the line XZ\ but there is evidently an infinite choice 
in the location of this point of intersection. 

Let WX, CX, be the lines of action of one pair of possible 
forces, P.^j acting at l^and Pc at C. Lay off XY =^ Q, and find 
its components, AX = P.^ and BX= P^. Now, through A 
draw AD parallel to WC, making two force triangles AXD, 
YAD. In AXD, P^ is resolved into the component AD parallel 
to WC d.nd DXdXong Q. Pc is likewise resolved into DA and YD. 



[} 



'"M 



^zr 



"-^ 




232 FORCES ACTING IN AND UPON THE CONNECTING ROD. 

A similar resolution of P^, at W, by the force parallelogram 
WD'A'B\ and of Pc at C by CD"A"B", gives the same com- 
ponents as the figure XADY\ and, what is more, it gives the 
components on their lines of action, or locates them. For, since 
the forces P^, Pc act upon the rigid bar WC "dX the points J^and 
C, their components, by whatever resolutions, must also act at 
these points. Considering now the components of P^, Pc we 
have WD (= DX) and CD" [= YD), forming with Q a system 
of three parallel forces in equilibrium ; we will call these the 
equilibrium components , and designate them P^^, Pee. Besides 
these there are WB' = AD = P^, and CB" = DA = P^, which 
we will call the internal stress components, and which simply 
balance each other as they act along the line WC. 

3. Since the equilibrium components are taken parallel to Q, 
they must be invariable as long as they act at W and C, no 
matter what two coincident P forces are applied at these points. 
From the laws of parallel forces w^e would have 

P..e '\- Pce= Q and P^, • WZ = Pee ' CZ. 

This latter relation may also be demonstrated from Fig. 126: 



From triangles AXD, WXZ, 
XD XZ 



DA WZ' 



and from triangles A YD, CXZ, 
AD CZ 



DY XZ 



Whence JJ = ^> ^r P^e : Pee :: CZ : WZ 

4. Since every possible P.^ must have WD' as one of its com- 
ponents—the other component being along WC — it is evident 
that a line drawn through D' parallel to WC, as LM, Fig. 127, is 
a locus of the end of P^ when this is laid off from fFas an origin. 
Similarly, DM' is the locus of the end of P^. Fig. 127 illustrates 
the properties of these loci, and shows clearly how the internal 
stress components vary as P^ and Pc change. 

5. The validity of the preceding discussion depends upon the 
right to select some particular point on the line of action of a 
orce as its point of application, through which point all the com- 



CONSTRUCTIO.^ FOR THE PIN PRESSURES. 233 

ponents of the force must pass. Now forces acting upon a rigid 
body, as MN Fig. 128, can act and react upon each other only- 
through the medium of stresses developed in the body-stresses 
which are distributed over its cross-section, but whose effect in 
any direction must be represented by a single resultant force. To 
select a particular point of application is simply to make that the 
point where the external force meets the resultant of the internal 
stresses ; and obviously all components of the external force must 
pass through this point. But in the general case of Fig. 128, no 
reason can be assigned for the selection of any particular point 
on, for instance, the line AB; not even that the point must fall 
within the body. The total effect, along the line AE, of the 
internal stresses will not vary as the point moves ; but their com- 
ponent effect, in any assigned direction, will vary with the same 
component of the external force, equilibrium being always pre- 
served. 

It will be well to realize that, when we make the force resolu- 
tion of Fig. 126, as at EGKF^ Fig. 128, and call AB the Hne of 
internal stress, that only a part of the internal stress effect acts 
along AB ; there are also developed stresses, closely analogous 
to those in a beam, under the action of the parallel forces Q, P^e, 

Pee- 

Fig. 128 is intended to illustrate choice of hne of internal stress 
(for the force resolution of Section 2), which exists in the general 
case of any rigid body in equilibrium under the action of three 
forces, showing two out of an infinite number of possible cases. 
It is evident that we m.ay impose some condition, as for instance 
Pyje shall act along the line WD, which will at once determine 
the point of application ci P-^. 

Construction for the Pin Pressures, when Friction 
IS Neglected. 

6. In Fig. 129, the resultant of weight and inertia of rod is sup- 
posed to be known, and is represented by XY = Q. The pin- 
pressures must pass through the centers of the pins ; the force 



234 CONSTRUCTION FOR THE PIN PRESSURES. 

diagram for determining them will be constructed at IV; for P^, 
besides balancing its share of Q, must also be the resultant of the 
known effective horizontal pressure on the wrist-pin and of the 
guide reaction whose direction is known ; these two conditions 
determine it fully. 

Lay off WJV, parallel, equal, and opposite in direction to Q. 
Let it be divided at L, so that WL : LN :: CZ : ZW; then 
WL is the equilibrium component of P^. The line LM, parallel 
to JVC, is, in reference to the point W, a locus expressing graph- 
ically the first condition, as is shown in Section 4. Now measure 
off IVV = effective horizontal pressure, and draw Fil/ parallel 
to the guide reaction ; then Pi^ embodies the second condition; 
and the intersection of these two loci at M fixes WM as the 
actual pressure of wrist-pin on connecting-rod. The line MN, 
completing the triangle WMN, is the crank-pin pressure, whose 
line of action is CX. 



Pin Pressures when Friction is Considered. 

7. In Figs. 130 and 136, the circles around ^and (7 are the 
friction circles of the respective pins, greatly exaggerated. The 
mechanism is simply outlined. The pin pressures act tangent to 
these circles ; they must, of course, hold Q in equilibrium. But 
the lack of a point common to all the forces possible of action at 
each pin, and which may be taken as their point of application, 
makes the graphical expression of this condition much more 
difficult than in the preceding case. Fig. 130 illustrates the 
construction of an exact polar locus of the force P^/, the same 
in its conditions as LM, Fig. 127. It is obtained by drawing 
pairs of pin-pressure lines to various points on the line of Q as 
Xj, X^, X3, and resolving Q along these lines ; then laying off 
from W as a. pole the various P^Js thus found and drawing a 
curve DD through their ends. 

This curve has some interesting properties. If we draw WD 
parallel to Q, it corresponds to the case of pin-pressures parallel 
to Q. Note how WL differs from WL in Fig. 129 (the figures 



CONSTRUCTION FOR THE PIN PRESSURES. 235 

bein£^ alike, aside from friction). The branch DD^, to the right 
of D corresponds to the case of rod in compression. It reaches 
00 when the pin pressures are conceived as acting along the line 
EF^ tangent to the friction circles. For the left-hand branch 
DD^, rod in tension, the similar limiting case is the force direction 
KH. If the direction of motion of the mechanism were reversed, 
ihe entire locus would be changed. 

8. To establish an approximate construction of practical utility, 
we impose the condition that the equilibrium components of the 
pin-pressures, parallel to Q^ shall pass through the pin centers ; 
in Fig. 131 their lines of action are WL, CU. This makes them 
the same as in Fig. 129. Now draw from any point X on the 
line of Q the pin-pressure lines XR, XU\ the intersections R, U, 
become the points of application of pin pressures, through which 
all their components must pass, and RUis the line of the inter- 
nal stress components (see end of section 5). If we were to 
draw through W a. line parallel to ^Xand measure off the length 
of P^jr : then a parallel to RU, through L, would pass through the 
end of P«^(compare WM, LM, Fig. 129), and at that point it would 
agree with the locus DD oi Fig. 130; so that we have here 
stated, in a slightly different form, the method of obtaining a 
point on that locus. 

When the force Q of the rod is not taken into account, the 
internal stress line is ST, tangent to the friction circles, as stated 
in Section i. The approximation which we now make is to 
consider that this line 57" practically agrees in direction with the 
variable line RU, which in the actual rod is true within the limits 
of graphical accuracy. Then LM, parallel to ST^ becomes a 
locus exactly similar to LM of Fig. 129 or of Fig. 127. The 
pin-pressure construction is now the same as in Fig. 129, differ- 
ing only in that the guide reaction is inclined from the normal 
through the angle of friction. If the rod be in tension the locus 
is LM' parallel to a tangent drawn like KH of Fig. 1 30. Note 
that in Fig. 131 the actual lines of action of P^^/, Pc/ are not 
shown. 



236 THE ROD FORCE AND ITS COMPONENTS. 

Drawing this locus M'LMiw Fig. 130, we can get a good idea 
of the error of the approximate method. It is evidently greater 
when Q is large relatively to the force transmitted by the rod. The 
lines LM, LM' are asymptotes to the true locus; for -dX ^ RU 
agrees with ST and the construction is exact. In Fig. 1 3 1 the 
condition was stated that the pin- pressures must be tangent to 
the friction circles. To substitute ST for RU diS the line of 
internal stress is to make J9 and T the points of application of 
all pin-pressures found by the approximate loci, thus making 
these pressures differ slightly from those found by the exact 
construction. The approximation to tangency may be easily 
estimated by inspection. 



The Rod Force and its Components. 

9. Referring to Sections 6 and 8, it appears that, in order to 
construct the pin-pressures it is necessary to know the total rod 
force and its parallel components at the pin centers, the latter 
being a function of its location on the rod. In Fig. 132 is 
developed a method for determining these quantities. 

Refer to pp. 93-100, and foot note p. 44. 

In Fig. 132, WCO represents the mechanism. CO is taken to 

represent the acceleration — - of C\ and WO, equal acceleration 

R 

of W, is found as in Fig. 16, page 41 or by Eq. 36. Now 
by producing Cw to W, making CW equal CW, and completing 
triangle CW A similar to triangle CwO ; then revolving it into 
the position CWA, we find the center of acceleration in A. 

(For-^^= ^^, and angle AWO = angle A CO.) 
A W A C 

Cw, being the acceleration image of the rod, we have only to find 
center of gravity, g, of the image from G, and draw gO, to get 
the intensity and direction of the acceleration of the center of 
gravity G of the rod. Again, letting OC represent to a suitable 
scale of forces, the inertia (centrifugal) force which the mass of 



THE ROD FORCE AND ITS COMPONENTS. 237 

the rod would have if concentrated at C; then to the same scale, 
Og represents the actual inertia force of the rod. 

10. To locate this force we proceed as follows : 

The actual distributed mass of the rod may be replaced, for 
dynamic purposes, by two simple concentrated masses m^, m^, 
located on any straight line through the center of gravity, at the 
distances h^, h, from the latter; the conditions being — 

1 . ?;/i + m^ = M, (same mass). 

2. 7njij^ = mjt^, (same center of gravity). 

3. mji^ -\- inji"^ = Mk^, (same moment of inertia). 

k = principal polar radius of gyration, and k"" = hji^. 

The points of mass concentration chosen, and their accelera- 
tions found from the image, then the inertia forces of Mi, m^ will 
have the direction thus determined ; and through the intersec- 
tion of these inertia forces, components of the total inertia force 
of the rod, the latter force must pass. 

If we choose one mass point so that its inertia acts along the 
line WC\ then the direction of the inertia of the other mass point 
is of no importance, for it will intersect the first force in its own 
mass point, and the latter will serve to locate the resultant, or 
total inertia of rod. Construction : make angle WAB = angle 
CWO = a, so that angle ^^PF'= angle y^ 1^0; whence accelera- 
tion of B is along BC; at G erect GK perpendicular to rod, and 
equal to K; complete the right-angled triangle BK/, making 
BG X GJ = GK^^ then J is the second mass point, through 
which must pass the inertia force QE = Og. 

The weight and inertia being combined at Q, their resultant 
QF cuts the rod at P; and QF, divided in the ratio CP: PW is 
what we wish to obtain, but by a more compact and convenient 
construction. 

11. Lay off (70 = weight of rod, then Og = QF = total 
rod force Q. Construct on the image Cw a figure similar to 
CBKJ\ determine b by Ob parallel to CW\ lay off kg in proper 
ratio Cw, and find/; draw^y parallel to ivO, and J' e parallel to 

16 



238 THE ROD FORCE AND ITS COMPONENTS. 

Cb, then ge : eO ■= Cj : jw = CJ : JW. That is, the inertia 
force is divided into its parallel components at the pins. 

Now divide the weight 00' at Tso that O'T : TO :: CG : GW, 
separating the weight also into its parallel components at the pins. 
Starting from 0' , combine O'T and TR (= eg), the wrist-pin 
components of weight and of inertia. Their resultant 0' R is a 
possible wrist-pin pressure P^ (acting in the direction "rod on 
pin "); its end R must therefore lie on the P^ locus of which 0' 
is the pole. This locus is RS parallel to WC, and it cuts from 
6^^ the wrist-pin component 0' S. (Compare Figs. 126 and 127.) 
This completes the solution of the problem. 

Cw changes with the crank position, and with it kg. To avoid 
making repeated computations for the latter, we can make part of 
this image construction on the crank OC instead of on the accel- 
eration image Cw. Project b to B\ lay off G' K' in proper pro- 
portion to (9(7 and obtain y. The entire construction is now in- 
dependent of the rod, except for the direction of the lines Ob, RS. 

12. Fig. 133 shows the whole construction for one crank 
position, and will serve to illustrate the following instructions. 
Having necessary data, including weight, center of gravity, and 
radius of gyration of rod: 

F 

1. Draw circle ACD, with radius = -4- for rod, to a conve- 

A 

nient scale of forces, and space off C points. 

2. From p. 37 compute and lay out Ow ; draw images Cw. 

3. Strike circle HG, G being center of gravity of crank image 
of rod ; obtain g by projection ; lay off 0T0\ draw Og, O'g. 

4. Draw GKonce; then strike circle LK, and from each G 
locate the corresponding K on this circle. 

Om R 
^. Rod angle construction : — — - = — ; (9(7 cuts circle mpn at 
^ ^ OA L ^ 

p\ project / to 2/ ; ?/ (9 is parallel to rod, for crank position C 

(and for 180° — (7); keep this construction below the center 

line AO. 



THE ROD FORCE AND ITS COMPONENTS. 239 

6. Having b project to B, draw BKJ,Je. 

7. Lay off OE = eg, ER = 0T\ draw RS parallel to uO. 
To avoid confusion : 

{a) Complete each step in the above for all the crank positions 
before beginning the next. 

{b) Mark the points C, zv, G,g, 0\ T\ K, b, BJ, e, E, R, S, 
as found, by inki7ig a small circle about each, the first six and 
last three in black, the others in red. Letter one construction 
fully, and number each set of points with the degrees of 
crank angle, from o° to 360°; likewise, designate the rod line 
tiO with the angles to which they belong. 

[c) Draw as few lines as possible ; thus, on upper half of 
figure, CO need not come inside of circle HG\ below, of circle 
inu ; ub need not come to O ; projection lines, as Gg, bB, BK-KJ, 
Je, need not be drawn in. 

13. Fig. 133 completed, the pin-pressures are found by 
Figs. 134 and 135. Fig. 135 is drawn first. To as large 
a scale as paper will allow, strike arc AB, with rod-length L as 
radius; draw circle OC with, radius equal R\ put in crank posi 
tion OC', project C' to C\ WC is rod angle, and ST is line of 
internal stress, with friction. The play of forces, as affecting 
tangency of ST to friction circles, must be found from the 
diagram of effective horizontal pressures on wrist-pin ; modified 
at critical points by the inertia of the rod itself, as affecting 
direction of crank-pin pressure. It is possible to have one end 
of the rod in tension, the other in compression. Mark each ST 
with its crank angle. Fig. 134, is the same as the pin-pressure 
diagram of Fig. 131, except that all the forces are reversed, so 
as to give pressure of rod on pins rather than of pins on rod. 
Bunch the constructions in four groups on Plate 7, one for each 
quadrant of crank circle. NLW'is O'Sg to reduced scale. 



240 THE LINK MOTIONS. 



APPENDIX D. 

The Link Motions. 

In this set of valve gears, the eccentrics, the throw and location 
of each eccentric, are invariable. These Link Motions occur 
principally in the locomotive, which, to be sure, is a species of 
high-speed engine, but has so many other special problems con- 
nected with it that it cannot be fully treated within the limits of 
this work. Nevertheless, in order to give a certain degree of 
completeness to the general subject of valve-gear design we will 
here give an outline of the Kinematics of Link Motions. The 
Fink Motion, as used in the Porter-Allen Fngine, has already 
been discussed in the body of the text. 

In the valve gears known as Link Motions the same sort of 
steam distribution occurs as in the gears having a " Swinging 
Eccentric," that is, the locus of the eccentric centers is of the 
same general character. But the mechanism employed is very 
different, the adjustment of the Link Motions by hand causing 
the invariable eccentrics to produce the desired variations of valve 
motion. About the only case in which the adjustments are auto- 
matically made by the governor occurs in the Fink or Porter- 
Allen Motion. 

Dr. Zeuner found, in his well-known treatise on Valve- Gears, 
this locus of eccentric centers by analytical means, but it may be 
found much more easily graphically and, so far as practical ap- 
plications are concerned, with the same degree of accuracy. 

The basis of the graphical treatment is Fig. 66, which contains 
the method of finding the virtual eccentric for a valve motion 
whose stroke does not pass through the center of the shaft. 

We will here use the figure and demonstration given by Pro- 



THE LINK MOTIONS. 



241 



fessor A. Flies^ner in his work " Die UmsteuerunG^en der Loco- 
motiven.*" 

Fig. 136 represents what may be regarded as the general case 
of all the link motions, extensively used in practice, which impart 




Fig. 136. 

to the valve a motion like that which it experiences when directly 
driven by a single, simple, eccentric. This figure and Fig. 137 
show how two eccentric motions may be combined so as to be 
equivalent to the motion of a single eccentric, called the virtual 
eccentric. This, indeed, is the gist of almost all the link-motion 
problems. 

A'B' is regarded as the link provided with slots at the ends to 
permxit the eccentric ends ^^ and B' to move in the perfectly straight 
lines, g^ and g^. The point C on this link directly drives the 
valve and is itself assumed to move in a straight line g, which is 
parallel to the lines g^ and g^. The rods are of unequal lengths, 
/i and 4; the eccentricities p^ and p^ are also unequal and have 
the unequal angles of advance ^i and ^2- The position shown in 
full lines corresponds to crank position at left dead point ; the 



* Dr. Burmester has also given an elegant kinematic presentation in his 
" Lehrbuch der Kinematik," pp. 696-702, from which some of the special 
cases can be more rigorously deduced. 



242 



THE LINK MOTIONS. 



broken lines of the figure correspond to the position of the crank 
at the right-hand dead point. 

Here it will be well to note the difference between link-motions 
with open and with crossed rods. When crank is at dead point 
and eccentrics are placed between engine shaft-center and Hnk 
A' B' , then the mechanism is said to have crossed rods if the ec- 
centric rods cross in this position and open rods if the eccentric 
rods do not then cross. Fig 136 is an example of the open-rods 
variety. 




Take A and B as the centers of travel of points A and B' and 
join A and B ; then will AB pass through the intersection 5 and 
be the reference line from which the travel CO of any point C 



THE LINK MOTIONS. 243 

on link A B' can be estimated for any crank position. It is evi- 
dent from the figure that the travel CO at any instant is 

CO = AA' + -^^^ iBB' — AA'). 

Let us take these eccentrics 0K\, OK'^ separately and on a 
larger scale construct in Fig. 137, after the method of Fig. 66, 

the virtual eccentrics OK^^ and OK^, whose distances \ jJ ^ 
from the vertical line V are respectively equal to the travels 
{ 11' of point i'}'"''^ P'g- ■ 36. To do this we make 

K\ K, =j'X 0K\ and K\ K, = — ^ x 0K\ 
for the case of open rods, 

and K\ k,=K\K,=^ — ^ y. '0K\ 



and K\k^ = K'^K^ = -\- -f- X OK'., for crossed rods. 

Now join Ki and K^ and divide at K the distance K^^K^ so that 
the ratio 

K^K ^ j 

K,K q 

in other words, in the same ratio that the valve-stroke-line gy 
(Fig. 136) divides the total distance between g^ and ^2, the lines 
of travel of the eccentric pins. 

This will make the distance KS of point K from the vertical 
6>F equal to 

KS = K,S, + (KS, — K,S^ ^J— = AA'^ iBB'— AA)-J—. 

J -^q J ^ q 

When the crank turns into some other position, the distance 
between pins A' and B' in Fig. 136 is still divided into the same 

ratio -— — by the line of stroke^ of the valve, and in Fig. 1^7 

the triangle OKr KK^ goes into some other position OE^^ BE^^ 



244 THE LINK MOTIONS. 

where the distances of E^ and E,^ from 6>Fequal the corresponding 
distances of pins A and B' from their centers of motion A and^, 
and the distance ^ from (9Fis equal to the distance of the driv- 
ing point C of the valve from its center of motion C. The 
complicated motion with which we started has therefore been 
reduced to the simple motion produced by a virtual eccentric 
OK = OE. 

We will now show the applicability of this method to Stephen- 
son's, Gooch's and Allan's link motions and will assume the 
reader to be acquainted with the general kinematic features of 
these mechanisms. In each of these motions the distance be- 
tween the lines g-^ and ^2 (Fig- 1 36) remains unchanged. 

In Stephenson's motion the block or driving point O keeps its 
line of stroke ^, but the pins A' and B' change their lines of 
stroke ^i and ^2 to some other parallel position in such a way 
that the distance Cj -f c^, between them remains the same, while 

the ratio — ' is changed. As Cj, and c^ change the virtual eccen- 

€■2 

tries OK^ and OK^ will change; the size and position of OK 
depend on this change and also on the ratio of 7" to ^. The 
locus of K for different positions of the link we will construct 
later on. 

In Gooch's motion the link is stationary, the pins A and B' 
keeping their lines of stroke g^ and g^. The locus of the center 
K of the virtual eccentric OK will therefore for this mechanism 
always be the line K^K^ because c^ and c^ do not change. The 

position of K oxi this hne will depend on the ratio — . 

In Allan's motion both link and block are shifted and in oppo- 
site directions. Each of the three lines, ^i, g and g^ changes its 
place, the values of c^, C2,j'ind q change but so that <:'i + <^2 = 
J -\- q =z constant. The locus of the center K of the virtual 
eccentric depends on c., c^yj and q and is found as before. 

The Fink or Porter-Allen link motion may be regarded as 
that particular case of the Gooch in which the two eccentrics 



THE LIMK MOTIONS. 245 

are equal and have an angle of advance of 90° causing them to 
coincide and thus reducing it to a link motion with only one 
eccentric. (The link itself then becomes part of the eccentric 
strap and may be regarded as a bent lever rocking on a vibrating 
fulcrum). In the actual link motion this would have the effect 
of causing the two eccentric rods and the link to form one rigid 
piece. This would prevent any angular motion between either 
rod and the link and, in the Fink, the link would therefore receive 
the whole of the rocking motion due to the throw of the single 
eccentric. The result is to make the position of the locus of the 
centers ^of the virtual eccentric in Fink just what it should be 
if it were strictly a special case of the Gooch, but the location of 
K on this locus cannot be found without adding still another 
approximation to the series, namely, to substitute for length I 
of the eccentric rod the average distance A of the trunnion of 
link from the center of either eccentric. This substitute A will 
generally be shorter than X and gives a greater throw by making 

ordinate -iP^jP- The result will then agree with that found by 

analytical investigation.* 

In constructing the locus of the centers of the virtual eccentric 
and the diagrams of the four grades of expansion for forward 
motion we will use the examples of the Gooch, Stephenson and 
Allan gears taken by Dr. Zeuner for the same purpose in his 
Valve Gears, 

Gooch's Link Motion. Dimensions r= 2^'\ o = 20°, 2c = 
c^ -\- c^ = I2'^ / = 48'', e = 0.91'' and i = 0.23''. Let the 



* This is a doubtful derivation of the Fink throw from that of the Gooch, 
for which only the writer is responsible Dr. Burmester preferred to obtain 
it independently. Below we have given a simple analytical proof of the 
travel of the valve in the Porter-Allen link motion. 



246 



THE LINK MOTIONS. 



four grades of expansion correspond to the four cases in which 
block {i.e., line g) is at %c, yic, Y^c and c from the line of centers 




Fig. 138. 



OX(Fig. 136). In Fig. 138 c,.=^c,= c=:&\j= i/q and K,K/ = 



0.297 = K2KJ. The value of the ratio 



J 



= ^, ^, >^ando.. 



7 + ^ 

Make the offsets K,K^' = K^KJ =r 0.3'', join K^ and K^ then 
divide K^E^^ into four equal parts and draw OEv^, OE^^, ^-^^z^ 
and OKt_. The link K^K^ is the locus of the centers of the 
virtual eccentric for forward and backward motion and OE^^^ 
OEj,^, OE^^ and OK-^ represent in length and position the virtual 
eccentric themselves for the forward motion when the crank is 
at the dead point. The diagram thus found agrees exactly with 
that obtained from Dr. Zeuner's approximate formula. The 
diagram shows that lead nX constant. The locus for crossed rods 
is shown at k^k^. 



THE LINK MOTIONS. 247 

Porter-Allen Link Motion. The relation between the 
Fink and the Gooch motion was pointed out above, p. 245. If 
the angle of advance of each eccentric is made equal to 90°, then 
the two eccentrics OK^ and OKJ reduce to one (9 A" and the locus 
K^K^ of the virtual eccentrics takes the position K' KK" , For 
the reason already given, p. 246, the offset KK' = KK' for full 

grade should be greater than K^K\ = K^K^ = j P' According 

A 

to Zeuner it should be -7 ^ for the same grade, where A is the 

distance from center of eccentric to center of trunnion ; (this trun- 
nion in Porter- Allen gear joins sustaining arm and link on the 
center line of the slot). For any other grade of expansion Zeuner 

u 
gives offset = -7 p, where u is the perpendicular distance of block 

above line of centers OX. 

When — <4 we should plot positions of block to ascertain the 

valve travel but when — > 4 we may for the Fink gear, as real- 
ized in the Porter-Allen link motion, determine the travel ^ as 
follows : Consider the total movement of link to be composed 
of two parts, a horizontal motion parallel to itself which is due 
the horizontal throw of the eccentric and a rocking motion about 
the trunnion (or vibrating fulcrum) which is due to the vertical 
throw of the eccentric. These two motions together cause the 
block to move a distance c equal to the sum 

f) cos CO -{- -: p sm 0) = q. 

Here the first term is the horizontal throw common to all points 
of the link and the second term is the vertical throw p sin w of 

zi 
the eccentric X the ratio -: of the arms of the bent lever consti- 

A 

tuting the link and rotating about the trunnion. 



248 THE LINK MOTIONS. 

The formula just given is the polar equation of the valve circle 

u 
and the coordinates and — p of the vertex of its diameter show 

A 
that this vertex lies on a rectilinear locus perpendicular to the 
line of centers and at a distance p from the center of the shaft, 
which agrees with the result so irregularly deduced above. 



Allan Link Motion. Dr. Zeuner gives the following dimen- 
sions of an existing valve gear with crossed rods. 
Eccentricity, p =-. 2.75''. 

Angle of advance, d= 30°. 

Length of each eccentric rod / = 49. 2''. 

Half length of link, c = 6". 

Length of the radius rod, l^ z= 60". 



Distance of trunnion of hanger . , „ 

of radius rod from valve stem 



:em J 



Short lifting arm of the lever, a^ = 2.88'' 1 

/" ratio rzzr '^ '2'7 

Long Hfting arm of the lever, d^ = 6.81'' j 
Outside lap, e = W^. 

Inside lap, z = xV, 

Ratio 7- of block to link movement, — - = - X — = = 2.9. 

Ratio of distance u of block from dead point of link to distance 
2/1 that link is shifted from line of centers is ^^ = 3.9. 

For the four grades of expansion, u =c u= )^c\u = }4c u=}4c u—o 

Corresp'd'g amount of link shifting^ 0.256^ o.i92<r 0.128^: 0.064c 0.0 

Since Ci= c — ^i and c^^= c ^ Ut_ approximately, 

— p are 0.249 0.270 0.292 0.314 0.335 



the correspond- 
ing values of ^ 



^> are 0.420 0.399 0-378 0.356 0.335 



J 

— are o 



THE LINK MOTIONS. 249 

Lay off K^k^ = 0.25 on a line perpendicular to OK^', also 
KJk^ = 0.25 on a line perpendicular to OKJ; the points k^ and 
k^ will be points on the locus corresponding to ?c = c, or full 
gear forward and backward respectively. On each of these two 
perpendiculars lay off from K^ and K^, in the direction K^k^, 
and KJk^ the distance 0.335, and join the ends of these distances; 
where this connecting line cuts OC will be the points/ of the 
locus corresponding to mid-gear. This gives us three points kj_, 




Fig. 139, 

/ and k^ on the desired locus ; a circle through these three would 
closely represent the curve but the arc is so very flat that its 
center would fall off the drawing. We will, therefore, construct 
the arc. The lines are so close together that only the construction 
for half gear forward is shown. We make the offset at K^ equal 
to 0.292 and that at KJ equal to 0.378 and join the ends of these 



2^0 THE LINK MOTIONS. 

offsets. This connecting line is divided at D^^ in the ratio of i 
to 3, and therefore D^^ is another point on the locus. The other 
points are found in like manner. This locus can be shown to be 
a parabola and for the present case of crossed rods is convex to 
the center 0. 

For open rods the locus is K^P for forward gear and is again 
a parabola but is not exactly like locus k^p for forward gear and 
is again a parabola but is not exactly like locus k^p for crossed 
rods. We have however TP =z Tp and K/K^ = X//^i. The locus 
for open arms is concave to center 0. To find it we will use, 
without proof, a simpler demonstration given by Burmester. On 
the diagonal J(^^P, joining the known points X^ and P, construct 
the rectangle K^QPS\ draw the second diagonal QS. To find 
points of the locus on the parallels through ji, JF2 and y^ we 
draw through these points parallel to QS to their intersection 
with K^Q at q^, q^ and q^ respectively. The intersections of the 
lines Pq^, Pq^ and Pq^ with the corresponding parallels will be 
points on the desired locus. 

The valve-circles have only been drawn for the case of crossed 
rods, forward gear. They show but little variation in the lead. 
The corresponding eccentric positions, when crank is at dead 
point, are shown on the other side of the figure. 

Stephenson's Link Motion. The dimensions taken by Dr. 
Zeuner for his diagram : p = 2.36'^ S = 30°, X = 55'', c =z 5.9'' 
e =: 0.94'' and i = 0.27''. We will consider the grades of 
expansion represented by 

u = c, u =: y^c, 21 = ^<;, u = y^c, U r= o 

for open rods. Since c^ ^=^ c — u and c^ ^=1 c -^- u we have 
the corresponding values of 

yP = o 0.064 0.127 0.191 0.254 

~ () — 0.508 0.444 0.381 0.317 0.254 

/ 



THE LINK MOTIONS. 



251 



K^ and K^ are points of the locus of the centers ot the virtual 
eccentrics which correspond to ?/ = + <: for full gear foward and 
to ?/ = — c for full gear backward. To find the other points on the 
locus, make the offsets /O3. K\U, K\U^ K\ K^ respectively equal 
to, 0.06, 0.13, 0.19, 0.25; also on the perpendicular /('^iT, make 




Fig. 140. 



the offsets K' s^ K\s^, K\s^, IC^K^ equal to the corresponding 
values 0.44, 0.38, 0.32, 0.25. Join s^ and 4, ^2 and 4, s^ and t^. 



K^ and ^^2. On sj:^ make t^E^ 



^^^, on sj^ make tiEv.^ = ^''- 



and on s^t^ make t^ Ev^ = y^ sj^. The points E^ , E^^, E^^ and 
the mid-point P of Kj^K^ are points on the desired locus. 0K\ is 
at once actual and virtual eccentric, while OE^ , OE^^, OEv^ and 



252 THE LINK MOTIONS. 

OP dSG. all virtual eccentrics corresponding to the given grades ol 
expansion for forward motion. 

The locus K\PK\ for open rods is concave towards center 0. 
The locus K\pK'^ for crossed rods is convex towards center 0. 
The two loci are exactly equal and symmetrical to the line K\K' ^. 
That for crossed rods can be found in the same manner as the 
one for open rods, but we have preferred to find it by a much 
simpler method given by Burmester,* omitting the proof 

In the lower part of Fig. 140 on the diagonal connecting the 
known points K'^ and/ we construct the rectangle pQK' ^T and 
then draw the second diagonal Tk^. To find points of the locus 
on any parallels jj/j, y, and ^3 we draw through points ^^1,^2 and 73 
parallels to Tk^ letting them cut QK' ^ in q^^ q^, q^ Then the 
intersections /^i with parallel j/^, oi pq^ with j/^, of/^3 with jj/3 will 
^\yr^ three points on the desired locus. Of course the locus for 
open rods might have been found in the same manner. 

Drawing the valve-circles on the corresponding diameters 
OD^^, OD^^, ODy,^, OD^^, OD^^ and the lap circles we get, in the 
usual manner, the steam distribution for the different grades of 
expansion. The varying lead, mn^, n7n^, Jim^y nm^ and nm^ is 
particularly noticeable. We will show later how this variation 
may be reduced for this link motion. It is however not so 
serious a defect as has been commonly beHeved. 

Porter-Allen Link Motion. Dimensions are taken from 
the II J^ X 20 engine shown in text, Figs. 79-82. Here ^ = \" ,A 
= 6", d z=: 90°, c=:6'\ e=iT^, z=Tt. Thc engine **runs over" and 
the expansion and exhaust valves are both of the positive direct 
type. Between the link and expansion valves there are reversing 
levers whose arms multiply the motion of the link four-thirds. 
The diagram should be constructed as if the valve were driven 
by a correspondingly enlarv^ed link-motion. There is also a 
lever between link and exhaust valve, but this does not reverse 

* This method and its demonstration are given in the " Lehrbuch der Kine- 
matik " p. 722. For another easy method see Fliegner's " Umsteuerungen 
der Locomotiven," p. 81. 



THE LINK MOTIONS. 253 

the motion, it reduces the travel in the ratio of 4^ to iitI. The 
exhaust valve does approximately follow the law of eccentric mo- 
tion but the steam valve does not do it as well, for the reversing is 
effected by a dent lever of such an angle as to effect a differ- 
ential motion like that of the wrist-plate of the Corliss engine. 
This valve motion has been satisfactorily discussed in the text, 
see Figs. 83-91 inclusive, and the accompanying notes. For the 
purpose of comparison however the graphical methods of the 
preceding link motion may be followed. Then for the grades of 
expansion : 



.; 



U ^ C U = Zy^c II =: )/ic U = }^C 

we have c^ z=ic^=^c and the corresponding values of 

o y% y^ Y^ % 

Then on a scale of — make the offset — —0 ^ -" . 

Adjustments of Link Motions so as to Equalize Motion. 

"The requirements of a perfectly equalized link motion are: 
perfect quality of cut-off, of exhaust closure, lead opening and 
maximum port opening, together with absence of block-slip, be- 
tween the forward and return stroke of the piston for every sus- 
pension of the link from full gear forward to full gear back. Such 
theoretical excellence is absolutely impossible with ordinary link 
motion. But good practical qualities may be obtained by sacri- 
ficing the non-essential to the essential points of the motion." 

Auchincloss, from whose well-known work these remarks are 
quoted, then proceeds to lay out the motion by making the leads 
equal at mid-gear and the cut-offs equal at half and full gear for- 
ward and back, thus indicating that these elements constituted 
the essential points of the motion. At ^ cut-off however the 
condition favorable to minimum slip is introduced. The lead at 
full gear and the slip of the block then existing in the designed 
17 



2 54 THE LINK MOTIONS. 

link-motion are examined and the directions in which modifica- 
tions should be made are suggested * 

Zeuner, Burmester and Fliegner, on the other hand, seem to lay 
greater stress on equality of lead at the at ends for each 
grade of expansion and on the reduction of the slip. As these 
link motions are usually accompanied by long connecting rods 
the inequalities of cut-off" due to the angular motion of the rods 
is not very great and we believe them to be comparatively unim- 
portant, particularly at high speeds. We shall therefore follow 
the presentation given by Drs. Zeuner and Burmester. There 
are two leading principles or conditions: — 

1. A regular and correct distribution of the steam requires 
that the valve should move symmetrically on each side of a point 
(the center of motion) that remains fixed for all grades of expan- 
sion. 

2. A minimum slip of block requires that the trunnions of 
both the link and the radius rod should have as little vertical 
motion as possible, in other words, the chords of the arcs de- 
scribed by the lower ends of the hangers should be parallel to 
the central line of motion and the hangers themselves should 
be as long as possible. 

The first condition is satisfied in Gooch's motion by making 
radius of link-slot equal to length of the radius-rod; in Stephen- 
son's motion by making the arc of this slot equal to the length 
of the eccentric rod; and in Allan's motion by making the arms 

* To reduce the slip of the block if it should be excessive, Auchincloss 
recommends alterations in the following elements, (enumerating them in the 
order of relative efficiency), (i) Increasing the angle of advance, (2) Reducing 
the travel of the valve, (3) Increasing the length of the link, (4) Shortening 
the eccentric rods. When the proportions are unusual, as in link-motions 
for small launch-engines, considerable deviation from the standard arrange- 
ment is profitable, the irregularity introduced curing greater irregularities in 
the more important functions. In such cases it will always be well to check 
and correct work by a model. 



THE LINK MOTIONS. 255 

of the lever carrying the hangers (which move Hnk and radius- 
rod) bear to each other the ratio - = y ( i + aI i -f ^ j. In all 

three motions the effect is to give equal leads at the two ends 
for each grade of expansion, but the lead is not constant for 
all grades of expansion, except in the case of Gooch's 
motion. In Allan's motion the lead is slightly variable 

{ IkcreasinI }''''°'" f"" g^^*"^ to mid-gear in case of { °^^^^ } 

rods. In Stephenson's motion there is still greater variation and 
in the same direction as in Allan's gear. 

Dr. Zeuner has shown however that giving the eccentrics un- 
equal angles of advance would tend to correct this. A simple 
way of doing this is to change the angle between crank and 
eccentrics by moving back the crank when link drives the valve 
directly and by moving the crank forward (i. e. in direction of 
rotation) when there is a reversing lever between link and valve. 
The amount of this change in the setting is about 5°, or more 

exactly, it is the angle whose tangent is equal to y . When this 

modification is introduced the Stephenson becomes the simpler 
and more compact gear of the three and equally efficient. 

The second condition is not satisfactorily fulfilled in the case of 
the Allan motion as ordinarily constructed. The upper ends of the 
hangers should move in parabolic curves that are convex to each 
other. Instead of that they are guided in circular arcs that are con- 
cave to each other. In this case make the hangers as long as pos- 
sible, make total length of lever a-{ b=^l^ — ^ and give fulcrum of 
lever an abscissa (measured from center of shaft) equal to 

I -^ a and an ordinate equal to length of hanger. 

2/ 



256 THE LINK MOTIONS. 

In Gooch's motion the second condition is met by placing trun- 
nions of link at center of chord of the link-arc are utilized* and 

upper end of the hanger is placed so as to have an abscissa X — -— 

2/ 

(estimated from center of shaft) and an ordinate ^ = length of 
hanger. The abscissa is obtained by placing link at mid-gear 
with center of chord at points on central line of motion that 
correspond to the dead points of crank and then bisecting the 
distance between them. The condition is also fulfilled for the 
hanger of the radius-rod by finding the abscissa of point of 
suspension of the hanger in a similar way, namely, by placing 
crank at both dead points (for each grade of expansion) and 
the radius rod in corresponding positions (which will be nearly 
parallel) and then bisecting the horizontal distance between the 
two trunnion positions of the rod. The ordinate will be the 
length of the hanger and, erected perpendicular to the central 
line of motion at the point of bisection belonging to each 
grade of expansion, will give the locus of suspension for the 
uppper end of the hanger of the radius-rod. This locus is a 
circle with radius ^ struck from a center which is at the 
distance l^ above the mid-position of end of valve stem. 

In Stephenson's motion the second condition is only partially 
satisfied by the method of suspension usually employed in 
American practice. Concerning this matter of suspension the 
authorities differ according to their view of what constitutes an 
excellent gear. Dr. Zeuner's rule, of guiding the upper end of 
the hanger in the arc of a parabola, (having a parameter 2I and a 

vertex whose coordinates are X and ^) or in the arc of a 

2/ 

* Dr. Burmester chooses a different point than middle of chord for the 
trunnion. He finds that each point of suspension for hanger, on the vertical 

having the abscissa \ — — , has a particular trunnion point on the link-chord 

which will make the center of that chord travel most closely to the central 
line of motion. We do not reproduce the construction here, referring the 
reader to p. 711, Fig. 704, of Burmester's "Lehrbuch der Kinematik." 



THE LINK MOTIONS. 25/ 

circle with radius X^ , is probably best for maintaining equality of 
lead at each end of cyhnder. Dr. Burmester's way of plotting 
the locus of suspension is to place the driving point of the 
link corresponding to each grade of expansion on the central line 
of motion when the crank is at dead points, then join the 
trunnion points for these two link positions and bisect their 
distance apart by a perpendicular which of course passes through 
the point of suspension. This way of doing it makes the locus 
an / shaped curve which can be approximately replaced by a 
straight line and its effect will be to keep the slip of the block 
very small. On the other hand, Auchencloss's method of sus- 
pension keeps the cut-offs equal, the slip being minimized only 
for half cut-off; if in this case the slip should be regarded as 
excessive, the alterations mentioned in the foot note on p. 254 can 
be undertaken. 

The principal steps in the laying out of a Stephenson link 
motion are: 

I. Find position of center of rocker from mid-gear travel 
of link. 

II. Locate cut-off positions of rocker arm and find lap of 
valve. 

III. Find position of stud (or trunnion) on saddle for one-half 
cut-off, lor maximum cut-off and both for the forward and back- 
ward motion. The lines connecting the stud positions for the 
same cut-offs should be as nearly parallel as possible. 

IV. Locate the tumbling shaft for accompHshing an equalized 
cut-off in all positions ; (the tumbling shaft is the approximate 
center of the locus of suspension). 

V. Find and examine the lead at full gear for both strokes, 
also the extreme travel and the slip of the block. 

VI. Modification of the results by various expedients. 

All this work is graphical and is made with the help of a link 
template having on it the central arc of the link slot and notches 
representing the centers of the joints connecting link with eccentric 
rods. 



ONE COPY RECEIVED 
FEB 28 1903 



